Answer
$\frac{\partial f}{\partial x}(\frac{\pi}{4},\frac{\pi}{3})=-1$
$\frac{\partial f}{\partial y}(\frac{\pi}{4},\frac{\pi}{3})=\frac{1}{2}$
Work Step by Step
To solve these partial derivatives we will use chain rule.
The partial derivative with the respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(\cos(2x-y))=-\sin(2x-y)\frac{\partial}{\partial x}(2x-y)=-2\sin(2x-y),$$
because $y$ is treated as a constant.
The partial derivative with the respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(\cos(2x-y))=-\sin(2x-y)\frac{\partial }{\partial y}(2x-y)=-\sin(2x-y)\cdot(-1)=\sin(2x-y),$$
because now $x$ is treated as a constant.
Now we will evaluate these partial derivatives in the given point $(x,y)=(\frac{\pi}{4},\frac{\pi}{3})$:
$\frac{\partial f}{\partial x}(\frac{\pi}{4},\frac{\pi}{3})=\left.-2\sin(2x-y)\right|_{(x,y)=(\frac{\pi}{4},\frac{\pi}{3})}=-2\sin(2\cdot\frac{\pi}{4}-\frac{\pi}{3})=-2\sin\frac{\pi}{6}$
$\frac{\partial f}{\partial y}(\frac{\pi}{4},\frac{\pi}{3})=\left.\sin(2x-y)\right|_{(x,y)=(\frac{\pi}{4},\frac{\pi}{3})}=\sin(2\cdot \frac{\pi}{4}-\frac{\pi}{3})=\sin\frac{\pi}{6}$
$\sin\frac{\pi}{6}=\frac{1}{2}$
$\frac{\partial f}{\partial x}(\frac{\pi}{4},\frac{\pi}{3})=-2\sin\frac{\pi}{6}=(-2)(\frac{1}{2})=-1$
$\frac{\partial f}{\partial y}(\frac{\pi}{4},\frac{\pi}{3})=\sin\frac{\pi}{6}=\frac{1}{2}$