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.
 A: Usually, I would say use implicit differentiation.
But, if you know partial differentiation, we can say.
Let $z = x^4 + xy^3 + y^4 - 10$
And note that $z = 0$
Take the total derivative, and since $z$ is constant we know that the total derivative will equal $0.$
$\frac {dz}{dt} = \frac {\partial z}{\partial x}\frac {dx}{dt} + \frac {\partial z}{\partial y}\frac {dy}{dt} = 0$
$\frac {dy}{dx} = \dfrac {\frac {dy}{dt}}{\frac {dx}{dt}}$
$\frac {\partial z}{\partial x} + \frac {\partial z}{\partial y}\frac {dy}{dx} = 0\\
\frac {dy}{dx} = - \dfrac {\frac {\partial z}{\partial x}}{\frac {\partial z}{\partial y}}$
A: 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.
A: Simple: as the l.h.s. is constant, its total differential is $0$, i.e.
$$(8x^3+y^3)\mathrm d x+(3xy^2+4y^3)\mathrm d y=0,$$
whence, instantly,
$$\frac{\mathrm dy}{\mathrm dx}=-\frac{8x^3+y^3}{3xy^2+4y^3}.$$
A: $$2x^4+xy^3+y^4=10$$
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 
$\frac{\partial z }{\partial y}.$
A: 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:
$(8x^3+y^3)dx+(3xy^2+4y^3)dy=0\iff \dfrac{dy}{dx}=-\dfrac{8x^3+y^3}{3xy^2+4y^3}$
