We found that the solution to the exact differential equation is of the form

$f(x,y)=xy+(8e^x-8x{e}^x)-3x^3+h(y)$

Recall that $\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{y}}=N(x,y)=x.$ We have also solved the partial as $\frac{\text{d}\text{e}\text{l}\text{f}}{\text{d}\text{e}\text{l}\text{y}}=x+h^{\text{'}}(y).$ Therefore, $h^{\text{'}}(y)=0$ and $h(y)=c,$ for some constant.

To finish, find the general solution to the exact differential equation.