<s>[INST] 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.
 [/INST] $$2x^4+xy^3+y^4=10$$
D.w.r.t.$x$
we get $$ 8x^
3+y^3+3xy^2y'+4y^3 y'=0 \implies \frac{dy}{dx}=y'= -\frac{8x^3+y^3}{3xy^2+4y^3}$$
next you have actually found partial derivatives $\frac{\partial z }{\partial x}$ and 
$\frac{\partial z }{\partial y}.$
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