$$e^{x\sin y}$$
This requires the chain rule to differentiate with respect to $y.$ But why is this a composite function? Because to me it looks like two things multiplied together ($x$ and $\sin y$), which makes me want to use the product rule intuitively.
Clarification appreciated!
On
The function is in the form $e^z$ with $z=g(x,y)$ that is
$$f(x,y)=e^{g(x,y)}$$
therefore by chain rule
$$\frac{\partial}{\partial y}f(x,y)=e^{g(x,y)}\cdot \frac{\partial}{\partial y}g(x,y)$$
On
write $f(x,y)=x \sin y$ and $g(z)=e^z$
You are looking at $$h(x,y)=e^{x\sin y}=g(f(x,y))$$
$$\frac{\partial h}{\partial x}=g'(f(x,y)) \cdot \frac{\partial f}{\partial x}$$ $$\frac{\partial h}{\partial y}=g'(f(x,y)) \cdot \frac{\partial f}{\partial y}$$
You can finish the proof.
On
If $x$ and $y$ are both independent variables then you have $$ \frac \partial {\partial y} e^{x\sin y} = e^{x\sin y} \cdot x\cos y. $$ But if $x$ is a function of $y$ then you need both the chain rule and the product rule: $$ \frac d {dy} e^{x\sin y} = e^{x\sin y}\left( \frac{dx}{dy}\cdot\sin y + x\cos y \right). $$
In this context, $x$ is a constant. Therefore, using the product rule is an overkill.
On the other hand, your function is $\exp\circ f$, where $f(y)=x\sin(y)$. So, yes, it is a composite function.