Given $\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{=}\mathbf{0}$and $\mathit{g}\mathbf{(}\mathit{x}\mathbf{,}\mathit{y}\mathbf{,}\mathit{z}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{,}$find a formula for $\mathit{d}\mathit{y}\mathbf{/}\mathit{d}\mathit{x}$.

The given functions are given as:

$$\begin{gathered} f\left(x,y,z\right)=0 \\ g\left(x,y,z\right)=0 \end{gathered}$$

Partial differentiation is the method of determining a function's partial derivative. By holding the other variable constant, one can use this method to get the partial derivative of a function with respect to one variable.

Partial derivative formula:

$$\begin{gathered} {\mathit{f}}_{\mathbf{x}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{x}} \\ {\mathit{f}}_{\mathbf{x}}\mathbf{=}\underset{\mathbf{h}\mathbf{\to }\mathbf{0}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{h}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{-}\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}}{\mathbf{h}} \end{gathered}$$

Consider the given functions:

$f(x,y,z)=0$ …… (1)

$g(x,y,z)=0$ …… (2)

Differentiate equation (1) partially usethe chain rule:

$\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz=0$ …...(3)

Differentiate equation (2) partially use the chain rule:

$\frac{\partial g}{\partial x}dx+\frac{\partial g}{\partial y}dy+\frac{\partial g}{\partial z}dz=0$ …… (4)

Now, find the value of from equation (4) as shown below:

$\frac{\partial g}{\partial z}dz=-\left(\frac{\partial g}{\partial x}dx+\frac{\partial g}{\partial y}dy\right)$

$dz=-\frac{\left({\displaystyle \frac{\partial g}{\partial z}}dx+{\displaystyle \frac{\partial g}{\partial y}}dy\right)}{\frac{\partial g}{\partial z}}$ ……. (5)

Substitute the value of dz from equation (5) in equation (3) as follows:

$$\begin{gathered} \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy+\frac{\partial f}{\partial z}dz=0 \\ \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy-\frac{\partial f}{\partial z}\frac{\left({\displaystyle \frac{\partial g}{dx}}dx+{\displaystyle \frac{\partial g}{\partial y}}dy\right)}{{\displaystyle \frac{\partial g}{\partial z}}}=0 \\ \frac{\partial g}{\partial z}\frac{\partial f}{\partial x}dx+\frac{\partial g}{\partial z}\frac{\partial f}{\partial y}dy-\frac{\partial f}{\partial z}\frac{\partial g}{dx}dx-\frac{\partial f}{\partial z}\frac{\partial g}{\partial y}dy=0 \end{gathered}$$

Take common dy and dx in the above equation as follows:

$$\begin{gathered} \left[\frac{\partial g}{\partial z}\frac{\partial f}{\partial y}-\frac{\partial f}{\partial z}\frac{\partial g}{\partial y}\right]dy-\left[\frac{\partial f}{\partial z}\frac{\partial g}{\partial z}-\frac{\partial g}{\partial z}\frac{\partial f}{\partial x}\right]dx=0 \\ \left[\frac{\partial g}{\partial z}\frac{\partial f}{\partial y}-\frac{\partial f}{\partial z}\frac{\partial g}{\partial y}\right]dy-\left[\frac{\partial f}{\partial z}\frac{\partial g}{\partial z}-\frac{\partial g}{\partial z}\frac{\partial f}{\partial x}\right]dx \end{gathered}$$

Thus, a formula for $\frac{dy}{dx}$is given byrole="math" localid="1659152839895" $\frac{dy}{dx}=\frac{\left[\frac{\partial f}{\partial z}\frac{\partial g}{\partial z}-\frac{\partial g}{\partial z}\frac{\partial f}{\partial x}\right]}{\left[\frac{\partial g}{\partial z}\frac{\partial f}{\partial y}-\frac{\partial f}{\partial z}\frac{\partial g}{\partial y}\right]}$ .