$$$0\=f_x x_y\+f_y$ ⇒ $x_y\= -\frac{f_y}{ f_x}$$$

and

$1\=f_x x_z\+f_z$ ⇒ $z_x\= \frac{1- f_z}{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 4.3.20(a)   Implicit differentiation task template

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

$f\left(x\,y\,z\right)\=0$

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

$f\left(x\left(y\,z\right)\,y\,z\right)\=0$

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

${\mathrm{D}}_1\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)\left(\frac{\partial }{\partial y}x\left(y\,z\right)\right)\+{\mathrm{D}}_2\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)\=0$

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

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

Table 4.3.20(b)   Context Panel implementation of implicit derivative via first principles

Immediate results via the implicitdiff command

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

$\mathrm{implicitdiff}\left(z\=f\left(x\,y\,z\right)\,x\left(y\,z\right)\,z\right)$ = $-\frac{-1\+{\mathrm{D}}_3\left(f\right)\left(x\,y\,z\right)}{{\mathrm{D}}_1\left(f\right)\left(x\,y\,z\right)}$$$

Solution from first principles

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

$0\={\mathrm{D}}_1\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)\left(\frac{\partial }{\partial y}x\left(y\,z\right)\right)\+{\mathrm{D}}_2\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)$

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

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

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

$1\={\mathrm{D}}_1\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)\left(\frac{\partial }{\partial z}x\left(y\,z\right)\right)\+{\mathrm{D}}_3\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)$

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

$\frac{\partial }{\partial z}x\left(y\,z\right)\=-\frac{-1\+{\mathrm{D}}_3\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)}{{\mathrm{D}}_1\left(f\right)\left(x\left(y\,z\right)\,y\,z\right)}$