$$$2 x x_y\+2 y\=f_x x_y\+f_y$ ⇒ $x_y\= \frac{f_y-2 y}{2 x- f_x}$$$

and

$2 x x_z\=f_x x_z\+f_z$ ⇒ $z_x\= \frac{ f_z}{2 x- f_x}$$$

Obtaining $x_y$ via the Context Panel

Tools≻Tasks≻Browse:

Calculus - Differential≻Derivatives≻Implicit Differentiation≻$f\left(x\,y\,z\right)\=g\left(x\,y\,z\right)$ 

Table 4.3.22(a)   Implicit differentiation task template

$x^2\+y^2\=f\left(x\,y\,z\right)$

$x^2\+y^2\=f\left(x\,y\,z\right)$

$\stackrel{\text{evaluate at point}}{\to }$

${x\left(y\,z\right)}^2\+y^2\=f\left(x\left(y\,z\right)\,y\,z\right)$

$\stackrel{\text{differentiate w.r.t. y}}{\to }$

$2x\left(y\,z\right)\left(\frac{\partial }{\partial y}x\left(y\,z\right)\right)\+2y\={\mathrm{D}}_1\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)\left(\frac{\partial }{\partial y}x\left(y\,z\right)\right)\+{\mathrm{D}}_2\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)$

$\stackrel{\text{isolate for diff(x(y,z),y)}}{\to }$

$\frac{\partial }{\partial y}x\left(y\,z\right)\=\frac{-2y\+{\mathrm{D}}_2\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)}{2x\left(y\,z\right)-{\mathrm{D}}_1\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)}$

Table 4.3.22(b)   Context Panel implementation of implicit derivative via first principles

Immediate results via the implicitdiff command

$\mathrm{implicitdiff}\left(x^2\+y^2\=f\left(x\,y\,z\right)\,x\left(y\,z\right)\,y\right)$ = $-\frac{-2y\+{\mathrm{D}}_2\left(f\right)\left(x\,y\,z\right)}{{\mathrm{D}}_1\left(f\right)\left(x\,y\,z\right)-2x}$$$

$\mathrm{implicitdiff}\left(x^2\+y^2\=f\left(x\,y\,z\right)\,x\left(y\,z\right)\,z\right)$ = $-\frac{{\mathrm{D}}_3\left(f\right)\left(x\,y\,z\right)}{{\mathrm{D}}_1\left(f\right)\left(x\,y\,z\right)-2x}$$$

Solution from first principles

$\mathrm{temp}≔\mathrm{diff}\left(x{\left(y\,z\right)}^2\+y^2\=f\left(x\left(y\,z\right)\,y\,z\right)\,y\right)$

$2x\left(y\,z\right)\left(\frac{\partial }{\partial y}x\left(y\,z\right)\right)\+2y\={\mathrm{D}}_1\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)\left(\frac{\partial }{\partial y}x\left(y\,z\right)\right)\+{\mathrm{D}}_2\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)$

$\mathrm{isolate}\left(\mathrm{temp}\,\mathrm{diff}\left(x\left(y\,z\right)\,y\right)\right)$

$\frac{\partial }{\partial y}x\left(y\,z\right)\=\frac{-2y\+{\mathrm{D}}_2\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)}{2x\left(y\,z\right)-{\mathrm{D}}_1\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)}$

$\mathrm{temp}≔\mathrm{diff}\left(x{\left(y\,z\right)}^2\+y^2\=f\left(x\left(y\,z\right)\,y\,z\right)\,z\right)$

$2x\left(y\,z\right)\left(\frac{\partial }{\partial z}x\left(y\,z\right)\right)\={\mathrm{D}}_1\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)\left(\frac{\partial }{\partial z}x\left(y\,z\right)\right)\+{\mathrm{D}}_3\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)$

$\mathrm{isolate}\left(\mathrm{temp}\,\mathrm{diff}\left(x\left(y\,z\right)\,z\right)\right)$

$\frac{\partial }{\partial z}x\left(y\,z\right)\=\frac{{\mathrm{D}}_3\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)}{2x\left(y\,z\right)-{\mathrm{D}}_1\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)}$