find all the points on the graph x^4+y^4+2=4xy^3
Your question as asked does not make sense...
Do you mean graph it?
Since the title mentioned a derivative we can use implicit differentation to find it.
$\displaystyle 4x^3+4y^3\frac{dy}{dx}=4y^3+12xy^2\frac{dy}{dx}$
From here you can solve for $\displaystyle \frac{dy}{dx}$
Ahhhh so now we can finish
So we know at the horizontal tangents that the derivative is zero
$\displaystyle \frac{dy}{dx}=0$
Now using the equation from above we get
$\displaystyle 4x^3+4y^3\frac{dy}{dx}=4y^3+12xy^2\frac{dy}{dx}$
$\displaystyle 4x^3=4y^3 \iff y=x$
Now if we plug this back into the original equation we get
$\displaystyle x^4+y^4+2=4xy^3$ setting $\displaystyle y=x$ we get
$\displaystyle x^4+x^4+2=4x^4 \iff 0=2x^4-2 \iff 0=2(x^4-1)$
$\displaystyle 0=2(x^2-1)(x^2+1) \iff 0 =2(x-1)(x+1)(x^2+1)$
so it happens when $\displaystyle x=\pm 1$
So when x=1 we get
$\displaystyle 1+y^4+2=4y^3 \iff y^4-4y^3+3=0$
Now by the rational roots theorem we know the only possible rational roots are $\displaystyle \pm1, \pm 3$
If you check on y=1 works so (1,1) is a horizontal tangent
I will leave the other cases for you to check.
Note there may be other Real roots