<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] Using the product rule gives: $$d(f(u)g(v))=f'(u)g(v)du+f(u)g'(v)dv$$
Thus I like to work your equation like this:
$$d(2x^4+xy^3+y^4)=8x^3dx+y^3dx+3xy^2dy+4y^3dy=d(10)=0$$
Now we factorize to get the result:
$(8x^3+y^3)dx+(3xy^2+4y^3)dy=0\iff \dfrac{dy}{dx}=-\dfrac{8x^3+y^3}{3xy^2+4y^3}$
</s>