$$$z_x\=f_x\+f_z z_x$ ⇒ $z_x\= \frac{f_x}{1- f_z}$$$

and

$z_y\=f_y\+f_z z_y$ ⇒ $z_x\= \frac{f_y}{1- f_z}$$$

Obtaining $z_x$ 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 4.3.19(a)   Implicit differentiation task template

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

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

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

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

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

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

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

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

Table 4.3.19(b)   Context Panel implementation of implicit derivative via first principles

Immediate results via the implicitdiff command

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

$\mathrm{implicitdiff}\left(z\=f\left(x\,y\,z\right)\,z\left(x\,y\right)\,y\right)$ = $-\frac{{\mathrm{D}}_2\left(f\right)\left(x\,y\,z\right)}{{\mathrm{D}}_3\left(f\right)\left(x\,y\,z\right)-1}$$$

Solution from first principles

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

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

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

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

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

$\frac{\partial }{\partial y}z\left(x\,y\right)\={\mathrm{D}}_2\left(f\right)\left(x\,y\,z\left(x\,y\right)\right)\+{\mathrm{D}}_3\left(f\right)\left(x\,y\,z\left(x\,y\right)\right)\left(\frac{\partial }{\partial y}z\left(x\,y\right)\right)$

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

$\frac{\partial }{\partial y}z\left(x\,y\right)\=\frac{{\mathrm{D}}_2\left(f\right)\left(x\,y\,z\left(x\,y\right)\right)}{1-{\mathrm{D}}_3\left(f\right)\left(x\,y\,z\left(x\,y\right)\right)}$