$h_{\mathrm{xx}}\=\{\begin{array}{cc}-\frac{4x{y}^3\left(x^2-3y^2\right)}{{\left(x^2\+y^2\right)}^3}& \left(x\,y\right)\ne \(0\,0\\ 0& \left(x\,y\right)\=\left(0\,0\right)\end{array}$

$h_{\mathrm{yy}}\=\{\begin{array}{cc}-\frac{4x^{3}y\left(3x^2-y^2\right)}{{\left(x^2\+y^2\right)}^3}& \left(x\,y\right)\ne \left(0\,0\right)\\ 0& \left(x\,y\right)\=\left(0\,0\right)\end{array}$

$h_{\mathrm{xy}}\=\{\begin{array}{cc}\frac{x^6\+9x^4{y}^2-9x^2{y}^4-y^6}{{\left(x^2\+y^2\right)}^3}& \left(x\,y\right)\ne \left(0\,0\right)\\ -1& \left(x\,y\right)\=\left(0\,0\right)\end{array}$

$h_{\mathrm{yx}}\=\{\begin{array}{cc}{h}_{\mathrm{xy}}\left(x\,y\right)& \left(x\,y\right)\ne \left(0\,0\right)\\ 1& \left(x\,y\right)\=\left(0\,0\right)\end{array}$

$h_{\mathrm{xx}}\left(0\,0\right)\=\underset{h\to 0}{lim}\frac{h_x\left(h\,0\right)-0}{h}\=\underset{h\to 0}{lim}\frac{0-0}{h}\=0$

$h_{\mathrm{yy}}\left(0\,0\right)\=\underset{k\to 0}{lim}\frac{h_y\left(0\,k\right)-0}{k}\=\underset{k\to 0}{lim}\frac{0-0}{k}\=0$

$h_{\mathrm{xy}}\left(0\,0\right)\=\underset{k\to 0}{lim}\frac{h_x\left(0\,k\right)-0}{k}\=\underset{k\to 0}{lim}\frac{-k}{k}\=-1$

$h_{\mathrm{yx}}\left(0\,0\right)\=\underset{h\to 0}{lim}\frac{h_y\left(h\,0\right)-0}{h}\=\underset{h\to 0}{lim}\frac{h}{h}\=1$

Initialize

$H\left(x\,y\right)\=\frac{xy\left(x^2-y^2\right)}{x^2\+y^2}$$\stackrel{\text{assign as function}}{\to }$$H$$$

Obtain the first partials $h_x$ and $h_y$ for $\left(x\,y\right)\ne \left(0\,0\right)$ 

${\mathrm{D}}_1\left(H\right)\left(x\,y\right)$ = $\frac{y\left(x^2-y^2\right)}{x^2\+y^2}\+\frac{2x^{2}y}{x^2\+y^2}-\frac{2x^{2}y\left(x^2-y^2\right)}{{\left(x^2\+y^2\right)}^2}$$\stackrel{\text{simplify}}{\=}$$\frac{y\left(x^4\+4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Temp}$

$\mathrm{Hx}\left(x\,y\right)\=\mathrm{Temp}$$\stackrel{\text{assign as function}}{\to }$$\mathrm{Hx}$

${\mathrm{D}}_2\left(H\right)\left(x\,y\right)$ = $\frac{x\left(x^2-y^2\right)}{x^2\+y^2}-\frac{2x{y}^2}{x^2\+y^2}-\frac{2x{y}^2\left(x^2-y^2\right)}{{\left(x^2\+y^2\right)}^2}$$\stackrel{\text{simplify}}{\=}$$\frac{x\left(x^4-4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Temp}$

$\mathrm{Hy}\left(x\,y\right)\=\mathrm{Temp}$$\stackrel{\text{assign as function}}{\to }$$\mathrm{Hy}$

Obtain the second partials $h_{\mathrm{xx}}\,h_{\mathrm{yy}}$ for $\left(x\,y\right)\ne \left(0\,0\right)$ 

${\mathrm{D}}_{1\,1}\left(H\right)\left(x\,y\right)$ = $\frac{6yx}{x^2\+y^2}-\frac{6y\left(x^2-y^2\right)x}{{\left(x^2\+y^2\right)}^2}-\frac{8x^{3}y}{{\left(x^2\+y^2\right)}^2}\+\frac{8x^{3}y\left(x^2-y^2\right)}{{\left(x^2\+y^2\right)}^3}$$\stackrel{\text{simplify}}{\=}$$-\frac{4x{y}^3\left(x^2-3y^2\right)}{{\left(x^2\+y^2\right)}^3}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Temp}$

$\mathrm{Hxx}\left(x\,y\right)\=\mathrm{Temp}$$\stackrel{\text{assign as function}}{\to }$$\mathrm{Hxx}$

${\mathrm{D}}_{2\,2}\left(H\right)\left(x\,y\right)$ = $-\frac{6xy}{x^2\+y^2}-\frac{6x\left(x^2-y^2\right)y}{{\left(x^2\+y^2\right)}^2}\+\frac{8x{y}^3}{{\left(x^2\+y^2\right)}^2}\+\frac{8x{y}^3\left(x^2-y^2\right)}{{\left(x^2\+y^2\right)}^3}$$\stackrel{\text{simplify}}{\=}$$-\frac{4x^{3}y\left(3x^2-y^2\right)}{{\left(x^2\+y^2\right)}^3}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Temp}$

$\mathrm{Hyy}\left(x\,y\right)\=\mathrm{Temp}$$\stackrel{\text{assign as function}}{\to }$$\mathrm{Hyy}$

Obtain the second partials $h_{\mathrm{xy}}\,h_{\mathrm{yx}}$ for $\left(x\,y\right)\ne \left(0\,0\right)$ 

${\mathrm{D}}_2\left(\mathrm{Hx}\right)\left(x\,y\right)$ = $\frac{x^4\+4x^2{y}^2-y^4}{{\left(x^2\+y^2\right)}^2}\+\frac{y\left(8x^{2}y-4y^3\right)}{{\left(x^2\+y^2\right)}^2}-\frac{4y^2\left(x^4\+4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^3}$$\stackrel{\text{simplify}}{\=}$$\frac{x^6\+9x^4{y}^2-9x^2{y}^4-y^6}{{\left(x^2\+y^2\right)}^3}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Hxy}$

${\mathrm{D}}_1\left(\mathrm{Hy}\right)\left(x\,y\right)$ = $\frac{x^4-4x^2{y}^2-y^4}{{\left(x^2\+y^2\right)}^2}\+\frac{x\left(4x^3-8x{y}^2\right)}{{\left(x^2\+y^2\right)}^2}-\frac{4x^2\left(x^4-4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^3}$$\stackrel{\text{simplify}}{\=}$$\frac{x^6\+9x^4{y}^2-9x^2{y}^4-y^6}{{\left(x^2\+y^2\right)}^3}$$\stackrel{\text{assign to a name}}{\to }$$\mathrm{Hyx}$

Obtain the second partials $h_{\mathrm{xx}}\,h_{\mathrm{yy}}$ at $\left(x\,y\right)\=\left(0\,0\right)$ 

$\underset{h\to 0}{lim}\frac{\mathrm{Hx}\left(h\,0\right)-0}{h}$ = $0$$$$$

$\underset{k\to 0}{lim}\frac{\mathrm{Hy}\left(0\,k\right)-0}{k}$ = $0$$$

Obtain the second partials $h_{\mathrm{xy}}\,h_{\mathrm{yx}}$ at $\left(x\,y\right)\=\left(0\,0\right)$ 

$\underset{k\to 0}{lim}\frac{\mathrm{Hx}\left(0\,k\right)-0}{k}$ = $-1$$$

$\underset{h\to 0}{lim}\frac{\mathrm{Hy}\left(h\,0\right)-0}{h}$ = $1$$$$$

Initialize

$H≔\frac{xy\left(x^2-y^2\right)}{x^2\+y^2}\:$

Obtain the first partials $h_x$ and $h_y$ for $\left(x\,y\right)\ne \left(0\,0\right)$ 

$\mathrm{H\_\_x}≔\mathrm{simplify}\left(\mathrm{diff}\left(H\,x\right)\right)$

$\mathrm{H\_\_x}≔\frac{y\left(x^4+4x^2{y}^2-y^4\right)}{{\left(x^2+y^2\right)}^2}$

$\mathrm{H\_\_y}≔\mathrm{simplify}\left(\mathrm{diff}\left(H\,y\right)\right)$

$\mathrm{H\_\_y}≔\frac{x\left(x^4-4x^2{y}^2-y^4\right)}{{\left(x^2+y^2\right)}^2}$

Obtain the second partials $h_{\mathrm{xx}}\,h_{\mathrm{yy}}\,h_{\mathrm{xy}}\,h_{\mathrm{yx}}$ for $\left(x\,y\right)\ne \left(0\,0\right)$ 

$\mathrm{H\_\_xx}≔\mathrm{simplify}\left(\mathrm{diff}\left(H_{}\,x\,x\right)\right)$

$\mathrm{H\_\_xx}≔-\frac{4x{y}^3\left(x^2-3y^2\right)}{{\left(x^2+y^2\right)}^3}$

$\mathrm{H\_\_yy}≔\mathrm{simplify}\left(\mathrm{diff}\left(H\,y\,y\right)\right)$

$\mathrm{H\_\_yy}≔\frac{-12x^{5}y+4x^3{y}^3}{{\left(x^2+y^2\right)}^3}$

$\mathrm{H\_\_xy}≔\mathrm{simplify}\left(\mathrm{diff}\left(\mathrm{H\_\_x}\,y\right)\right)$

$\mathrm{H\_\_xy}≔\frac{x^6+9x^4{y}^2-9x^2{y}^4-y^6}{{\left(x^2+y^2\right)}^3}$

$\mathrm{H\_\_yx}≔\mathrm{simplify}\left(\mathrm{diff}\left(\mathrm{H\_\_y}\,x\right)\right)$

$\mathrm{H\_\_yx}≔\frac{x^6+9x^4{y}^2-9x^2{y}^4-y^6}{{\left(x^2+y^2\right)}^3}$

Obtain the second partials $h_{\mathrm{xx}}\,h_{\mathrm{yy}}$ at $\left(x\,y\right)\=\left(0\,0\right)$ 

$\frac{\mathrm{eval}\left(\mathrm{H\_\_x}\,\left[x\=h\,y\=0\right]\right)}{h}$ = $0$$$

$\frac{\mathrm{eval}\left(\mathrm{H\_\_y}\,\left[x\=0\,h\=k\right]\right)}{k}$ = $0$$$

Obtain the second partials $h_{\mathrm{xy}}\,h_{\mathrm{yx}}$ at $\left(x\,y\right)\=\left(0\,0\right)$ 

$\frac{\mathrm{eval}\left(\mathrm{H\_\_x}\,\left[x\=0\,y\=k\right]\right)}{k}$ = $\mathrm{-1}$$$

$\frac{\mathrm{eval}\left(\mathrm{H\_\_y}\,\left[x\=h\,y\=0\right]\right)}{h}$ = $1$