First-order differential equation problem

dy/dx=4(y^4) What I did: dx/dy = 1/(4(y^4); x= the integralof 0.25(y^-4) = -1/12(y^-3) + c The answer is at the back of the book is 3(y^3)(4x+c) + 1 =0..There's a whole dozen of other differential equations I can't do so I must be making some very fundamental mistake. Any help would be very appreciated.

$\displaystyle \begin{align*}\frac{dy}{dx}&=4y^4\\ \frac{dy}{y^4}&=4dx\\ \int y^{-4}dy&=4\int dx\\ \frac{y^{-3}}{-3}&=4x+c \bigg\slash \cdot 3y^3 \\ -1&=3y^3(4x+c)\\ 3y^3(4x+c)+1&=0\end{align*}$

Thank you so much for the quick and clear response! :)