Suppose that the relation $\frac{x^2}{2} + \frac{y^2}{2} + \frac{z^2}{2} + xy + xz =\frac{7}{2}$ defines $z$ as a function of $x, y$ around the point $(1, 1, 1)$. Find $\frac{dz}{dy}$ at $(1, 1, 1)$. Implicit differentiation can be used.
I am not sure how to tackle this problem. All I know is that it has something to do with partial derivatives? Can someone please help me?
We want to find the rate at which $z$ changes when we fix $x$ and vary $y$. Note that this rate can still change depending on what $x$ is, but when computing this rate, we will treat $x$ as a constant.
Differentiate both sides of the equation with respect to $y$, which produces $$\begin{align*} \frac{\partial}{\partial y}\left(\frac{x^2}{2} + \frac{y^2}{2} + \frac{z^2}{2} + xy + xz\right) &=\;\;\frac{\partial}{\partial y}\left(\frac{7}{2}\right)\\\\ \quad\frac{\partial}{\partial y}\left(\frac{x^2}{2}\right) + \frac{\partial}{\partial y}\left(\frac{y^2}{2}\right) + \frac{\partial}{\partial y}\left(\frac{z^2}{2}\right) + \frac{\partial}{\partial y}\left(xy\right) + \frac{\partial}{\partial y}\left(xz\right) &=\;\;\frac{\partial}{\partial y}\left(\frac{7}{2}\right)\\\\ 0+y+z\frac{\partial z}{\partial y}+x+x\frac{\partial z}{\partial y} & =\;\; 0\\\\\\ \frac{\partial z}{\partial y}&=\;\;\frac{-x-y}{x+z} \end{align*}$$