Answer
$$\frac{\partial h(x,y)}{\partial x}=-2xe^{-(x^2+y^2)}$$
$$\frac{\partial h(x,y)}{\partial y}=-2ye^{-(x^2+y^2)}$$
Work Step by Step
We will use chain rule to find both partial derivatives.
The partial derivative with respect to $x$ is:
$$\frac{\partial h(x,y)}{\partial x}=\frac{\partial }{\partial x}(e^{-(x^2+y^2)})=e^{-(x^2+y^2)}\frac{\partial }{\partial x}(-(x^2+y^2))=e^{-(x^2+y^2)}\cdot(-2x)=-2xe^{-(x^2+y^2)}$$
Here $y$ is treated as a constant.
The partial derivative with respect to $y$ is:
$$\frac{\partial h(x,y)}{\partial y}=\frac{\partial }{\partial y}(e^{-(x^2+y^2)})=e^{-(x^2+y^2)}\frac{\partial }{\partial y}(-(x^2+y^2))=e^{-(x^2+y^2)}\cdot(-2y)=-2ye^{-(x^2+y^2)}$$
because now $x$ is treated as a constant.
