Partially tricky equation

Firstly, let's treat $\ y$ as a constant and use partial differentiation (hence the name of the problem): $\ f'(x+y) = f'(x) +6xy +3y^2 - 2y$ Now let x = 0 $f'(y) = f'(0) +3y^2 -2y$ Let y=1 $f'(1) = f'(0) +1 = 1 \Rightarrow\ f'(0) = 0$ $\Rightarrow\ f'(y) = 3y^2 -2y \therefore\ f(y) = y^3 -y^2 +C$ If we substitute x = 0 into the original equation we get $\ f(y) = f(0) +f(y) \therefore\ f(0) = 0 = C$ $\therefore\ f(10) = 10^3 - 10^2 = 900$