An A level question about partial differentiation 
The equation of a curve is $2x^4+xy^3+y^4=10$. Show that $$\frac{dy}{dx}=-\frac{8x^3+y^3}{3xy^2+4y^3}.$$ 

I understand that you are to work out:
\begin{align}
\frac{dz}{dx} = 8x^3 + y^3\\
\frac{dz}{dy} = 3xy^2 + 4y^3
\end{align}
and therefore, $\dfrac{dy}{dx}$.
My answer almost matches the requirement, but I don't understand how the negative symbol got there.

You don't need to solve for $\frac{dz}{dx}$ or $\frac{dz}{dx}$. 
We know $2x^4+xy^3+y^4=10$. I will take the derivative with respect to $x$ on both sides of the equation. So $\frac{d}{dx}(2x^4+xy^3+y^4)=\frac{d}{dx}10 \implies 8x^3+y^3+3xy^2\frac{dy}{dx}+4y^3\frac{dy}{dx}=0$. Trying to solving for $\frac{dy}{dx}$, we have $8x^3+y^3=\frac{dy}{dx}(-3x^2-4y^3)$. Then $\frac{8x^3+y^3}{(-3x^2-4y^3)}=\frac{dy}{dx}$, and so $-\frac{8x^3+y^3}{3xy^2+4y^3}=\frac{dy}{dx}$. Does that make any sense? I can go in depth with some steps if you want.
