Answer
$$\frac{\partial f(x,y)}{\partial x}=-2$$
$$\frac{\partial f(x,y)}{\partial y}=2$$
Work Step by Step
Before we find partial derivatives $f_x$ and $f_y$ we willl first transform function $f$ as:
$$f(x,y)=\int_x^y(2t+1)dt+\int_y^x(2t-1)dt=\int_x^y(2t+1)dt-\int_x^y(2t-1)dt=\int_x^y(2t+1-2t+1)dt=2\int_x^ydt=2\left.t\right|_x^y=2(y-x)=2y-2x$$
Now we can differentiate.
The partial derivate with respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(2y-2x)=2\frac{\partial }{\partial x}(y)-2\frac{\partial}{\partial x}(x)=2\cdot0-2\cdot1=-2$$
because $y$ is treated as a constant.
The partial derivate with respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(2y-2x)=2\frac{\partial }{\partial y}(y)-2\frac{\partial }{\partial y}(x)=2\cdot1-2\cdot0=2$$
because now is $x$ treated as a constant.
