1. ## Graph tangent equation

I have graph $x^4+y^4+2=4xy^3$ and I need to find tanget equation at $(1;1)$ (on a graph) point.

I can't seem to get what to do with $x$s and $y$s can't convey neither of those

Do I need to do partial fluxion for $x$ and then for $y$? (well... these ones: $Fx'=...$ and $Fy'=...$ )

If yes, what should I do next with those 2 functions?

2. For implicit differentiation or Partial Fluxion as you call it. You apply the following:

1) Take the derivative with respect to x on both sides
$\frac{d}{dx}(x^4+y^4+2) = \frac{d}{dx}(4xy^3)$

2)Any where you see a term with no y's in it, differentiate like normal
$4x^3 + \frac{d}{dx}y^4 = \frac{d}{dx}(4xy^3)$

3) Now in the terms with y's in them, we differentiate with respect to y but keep the term \frac{dy}{dx}. This is essentially the chain rule.
$4x^3 + \frac{dy}{dx}4y^3 = \frac{d}{dx}(4xy^3)$

4) Now in this case we'll have to apply the product rule to the right hand side of the equation
$4x^3 + \frac{dy}{dx}4y^3 = 4y^3 + \frac{dy}{dx}12xy^2$

Now we solve for $\frac{dy}{dx}$

$\frac{dy}{dx}(4y^3 - 12xy^2) = 4y^3 - 4x^3$
$\frac{dy}{dx} = \frac{4y^3 - 4x^3}{4y^3 - 12xy^2}$

Now we substitute in the point (1,1) and we get:
$\frac{dy}{dx} = \frac{4-4}{4-12} = 0$
So the slope of the tangent line at that point is 0.

3. Tanget equation is just $y=1$ since it's horizontal for $x$ axis?