So we have been given the fact that $ f(3,y-1) = (y-1)(y-3) + f(y,y-1)$ and the assumption is $f_x(x,y)$ is differentiable. I then rearanged to make $f(3,y-1)-f(y,y-1)=(y-1)(y-13). $ Then we need to show that:
$$\int_1^3\int_0^{x-1} f_x(x,y)\,\mathrm dy \,\mathrm dx = -\frac{4}{3}$$
It seems to me the first step is to change the order of the iterated integral to integrate $f_x$ w.r.t. $x$ first. Thus giving $f(x,y)$ and the integral being:
$$\int_0^{x-1}f(3,y)\,\mathrm dy - \int_0^{x-1}f(0,y)\,\mathrm dy $$
but this doesn't really help. I then tried to rearrange the limits of the $x$ and $y$ bounds but while closer that didn't seem to help me. The only other process I could think of would be the product of two integrals but I assume that's not possible as it would seem $f_x(x,y)$ isn't single variable. I don't really know what my next step should be.
\begin{align*} \int_{1}^{3}\int_{0}^{x-1}\dfrac{\partial f}{\partial x}(x,y)\,\mathrm dy\,\mathrm dx&=\int_{0}^{2}\int_{y+1}^{3}\dfrac{\partial f}{\partial x}(x,y)\,\mathrm dx\,\mathrm dy\\ &=\int_{0}^{2}(f(3,y)-f(y+1,y))\,\mathrm dy\\ &=\int_{0}^{2}y(y-2)\,\mathrm dy\\ &=\dfrac{1}{3}y^{3}-y^{2}\bigg|_{y=0}^{y=2}\\ &=-\dfrac{4}{3}. \end{align*}