Answer
$$\frac{\partial f(0,0)}{\partial x}=-1$$
$$\frac{\partial f(0,0)}{\partial y}=0$$
Work Step by Step
The partial derivative with the respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(e^{-x}\cos y)=\cos y\frac{\partial}{\partial x}(e^{-x})=\cos y\cdot e^{-x}\frac{\partial}{\partial x}(-x)=-e^{-x}\cos y$$
because $y$ is treated as a constant. In the last two steps we have used the chain rule to find the derivative of $e^{-x}.$
The partial derivative with the respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(e^{-x}\cos y)=e^{-x}\frac{\partial }{\partial y}(\cos y)=-e^{-x}\sin y$$
because now $x$ is treated as a constant.
Now we will evaluate the partial derivatives in the given point $(x,y)=(0,0)$:
$$\frac{\partial f(0,0)}{\partial x}=\left.-e^{-x}\cos y\right|_{(x,y)=(0,0)}=-e^{-0}\cos 0=-1\cdot1=-1$$
$$\frac{\partial f(0,0)}{\partial y}=\left.-e^{-x}\sin y\right|_{(x,y)=(0,0)}=-e^{-0}\sin0=-1\cdot0=0$$