$$\partial{z}/\partial{x}-\partial{z}/\partial{y} = x-y $$
so, finding integral curves of $$dx/1=dy/(-1)=dz/x-y$$ is the above eqaution's solution which is $$x^2-2xy-2z=c_{1}$$$$y^2-2xy-2z=c_{2}$$ therefore $F=(c_{1},c_{2})$
but the given solution is $z=(x-y)^2/4+f(x+y)$
ps: this problem is a sub part of finding higher order partial dif's solution
First constant of integration gives $$dx=-dy \implies x+y=c_1$$ Second constant is given by $$-dy=\frac {dz}{x-y}$$ $$-\int (x-y)dy=z+c_2$$ Substitute $x=c_1-y$ $$-\int (c_1-2y)dy=z+c_2$$ $$-c_1y+y^2-z=c_2$$ Substitute back $c_1=x+y$ $$xy+z=c_2$$ Therefore $$z=-xy+f(x+y)$$ You can write $-xy=\frac 14 (-(x+y)^2+(x-y)^2)$ $$\boxed {z(x,y)=\frac 14(x-y)^2+f(x+y)}$$