Answer
$$\frac{\partial}{\partial x}F(x,y)=\cos(e^x);$$
$$\frac{\partial}{\partial y}F(x,y)=-\cos(e^y).$$
Work Step by Step
Here we will use the fact that
$$\frac{d}{dx}\int_a^xf(t)dt=f(x)$$
if $a$ is a constant.
Regard $y$ as constant and differentiate with respect to $x$:
$$\frac{\partial}{\partial x}F(x,y)=\frac{\partial}{\partial x}\int_y^x\cos (e^t) dt=\cos(e^x).$$
Regard $x$ as constant and differentiate with respect to $y$:
$$\frac{\partial}{\partial y}F(x,y)=\frac{\partial}{\partial y}\int_y^x\cos (e^t) dt=\frac{\partial}{\partial y}\left(-\int_x^y\cos(e^t)dt\right)=-\frac{\partial}{\partial y}\left(\int_x^y\cos(e^t)dt\right)=-\cos(e^y).$$
