Answer
$$\frac{\partial f(x,y)}{\partial x}=1-x^2$$
$$\frac{\partial f(x,y)}{\partial y}=y^2-1$$
Work Step by Step
Before we find partial derivatives $f_x$ and $f_y$ we will first transform function $f$ as:
$$f(x,y)=\int_{x}^{y}(t^2-1)dt=\int_x^yt^2dt-\int_x^ydt=\left.\frac{t^3}{3}\right|_x^y-\left.t\right|_x^y=\frac{y^3}{3}-\frac{x^3}{3}-(y-x)=\frac{y^3}{3}-\frac{x^3}{3}-y+x$$
Now we can differentiate.
The partial derivate with respect to $x$ is:
$$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(\frac{y^3}{3}-\frac{x^3}{3}-y+x)=\frac{\partial}{\partial x}(\frac{y^3}{3})-\frac{1}{3}\frac{\partial }{\partial x}(x^3)-\frac{\partial }{\partial x}(y)+\frac{\partial }{\partial x}(x)=0-\frac{1}{3}\cdot3x^2-0+1=1-x^2$$
because $y$ is treated as a constant.
The partial derivative with respect to $y$ is:
$$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(\frac{y^3}{3}-\frac{x^3}{3}-y+x)=\frac{1}{3}\frac{\partial }{\partial y}(y^3)-\frac{\partial }{\partial y}(\frac{x^3}{3})-\frac{\partial }{\partial y}(y)+\frac{\partial }{\partial y}(x)=\frac{1}{3}3y^2-0-1+0=y^2-1$$
because now $x$ is treated as a constant.