Answer
\begin{aligned} \frac{\partial z }{\partial x} = \frac{e^x \sin(y+z)}{1-e^x \cos(y+z)}
\end{aligned}
\begin{aligned}
\frac{\partial z }{\partial y} = \frac{e^x \cos ( y+z)}{ 1-e^x \cos(y+z)}\\
\end{aligned}
Work Step by Step
Given $$z=e^x \sin(y+z)$$
$$ \Rightarrow e^x \sin(y+z)-z=0$$
by letting $$F(x,y,z)= e^x \sin(y+z)-z$$
So, we have
$$F_x(x,y,z)=\frac{\partial F(x,y,z)}{\partial x}= e^x \sin(y+z)\\
$$
$$F_y(x,y,z)=\frac{\partial F(x,y,z)}{\partial y}= e^x \cos(y+z)$$
$$F_z(x,y,z)=\frac{\partial F(x,y,z)}{\partial z}=e^x \cos(y+z)-1$$
Also, we get
\begin{aligned} \frac{\partial z }{\partial x}&=-\frac{F_{x}(x, y,z)}{F_{z}(x, y,z)} \\
&=-\frac{e^x \sin(y+z)}{e^x \cos(y+z)-1} \\
&= \frac{e^x \sin(y+z)}{1-e^x \cos(y+z)}
\end{aligned}
\begin{aligned}
\frac{\partial z }{\partial y}&=-\frac{F_{y}(x, y,z)}{F_{z}(x, y,z)} \\
&=- \frac{e^x \cos ( y+z)}{ e^x \cos(y+z)-1}\\
&= \frac{e^x \cos ( y+z)}{ 1-e^x \cos(y+z)}\\
\end{aligned}
