Answer
\[
\begin{array}{l}
\left.\frac{d z}{d x}\right|_{y=1}=8 \\
\left.\frac{d z}{d x}\right|_{x=-1}=-2
\end{array}
\]
Work Step by Step
\[
-1=x \text { and } 4 y^{2}+x^{2}=z
\]
Thus, now obtaining the intersection point of both curves:
\[
\begin{array}{l}
(-1)^{2}+4 y^{2}=z\\
1+4 y^{2}=z
\end{array}
\]
Differentiating:
\[
\frac{d}{d y}\left(1+4 y^{2}\right)=\frac{d z}{d y}
\]
\[
8 y= \frac{d z}{d y}
\]
At the point (-1,1,5)
\[
\begin{array}{l}
\left.\frac{d z}{d y}\right|_{y=1}=8(1) \\
\left.\frac{d z}{d y}\right|_{y=1}=8
\end{array}
\]
$[b]$ similarly, for the plane $1=y$
Obtaining the intersection point of both curves:
\[
\begin{array}{l}
4(1)^{2} +x^{2}=z\\
4+x^{2}=z
\end{array}
\]
Differentiating:
\[
\frac{d}{d x}\left(4+x^{2}\right)=\frac{d z}{d x}
\]
\[
2 x=\frac{d z}{d x}
\]
At the point (-1,1,5)
\[
\begin{array}{l}
\left.\frac{d z}{d x}\right|_{x=-1}=2(-1) \\
\left.\frac{d z}{d x}\right|_{x=-1}=-2
\end{array}
\]
