$h_x\left(x\,y\right)\=H_x\left(x\,y\right)\=\frac{y\left(x^4\+4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}$ 

$h_y\left(x\,y\right)\=H_y\left(x\,y\right)\=\frac{x\left(x^4-4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}$

$h_x\left(0\,0\right)\=\underset{h\to 0}{lim}\frac{H\left(0\+h\,0\right)-0}{h}\=\underset{h\to 0}{lim}0\=0$

$h_y\left(0\,0\right)\=\underset{k\to 0}{lim}\frac{H\left(0\,0\+k\right)-0}{k}\=\underset{k\to 0}{lim}0\=0$

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 partial derivatives for $\left(x\,y\right)\ne \left(0\,0\right)$ 

$\frac{\partial }{\partial x} H\left(x\,y\right)$ = $\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}$$$

$\frac{\partial }{\partial y} H\left(x\,y\right)$ = $\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}$

Obtain $h_x\left(0\,0\right)$ and $h_y\left(0\,0\right)$ 

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

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

Obtain $h_x\left(0\,0\right)$ and $h_y\left(0\,0\right)$ from first principles

$\frac{H\left(h\,0\right)}{h}$ = $0$$$

$\frac{H\left(0\,k\right)}{k}$ = $0$$$

Initialize

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

Obtain the first partial derivatives for $\left(x\,y\right)\ne \left(0\,0\right)$

$\mathrm{simplify}\left(\mathrm{diff}\left(H\left(x\,y\right)\,x\right)\right)$ = $\frac{y\left(x^4\+4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}$$$

$\mathrm{simplify}\left(\mathrm{diff}\left(H\left(x\,y\right)\,y\right)\right)$ = $\frac{x\left(x^4-4x^2{y}^2-y^4\right)}{{\left(x^2\+y^2\right)}^2}$$$

Obtain $h_x\left(0\,0\right)$ and $h_y\left(0\,0\right)$

$\frac{H\left(0\+h\,0\right)-0}{h}$ = $0$$$

$\frac{H\left(0\,0\+k\right)-0}{k}$ = $0$$$