Answer
$$-\frac{x e^{x y}+1}{x \cos (x z)}$$
Work Step by Step
Given $$ \quad e^{x y}+\sin (x z)+y=0$$
Consider $$F(x,y,z )=e^{x y}+\sin (x z)+y=0 $$
Then
\begin{align*}
F_{y}&=x e^{x y}+1\\
F_{z}&=x \cos (x z)
\end{align*}
Then
\begin{align*}
\frac{\partial z}{\partial y}&=-\frac{F_{y}}{F_{z}}\\
&=-\frac{x e^{x y}+1}{x \cos (x z)}
\end{align*}
