What is the correct answer to this calculus problem (not homework)

Factual Questions
1

I am trying to teach myself calculus. I am having trouble with problem 9 on this page. I can follow the explanation and I fully understand each step. My problem is I tried to do it another way and arrived at a different answer. I can’t see where I could have gone wrong.

This is what I did:
y/x^3 + x/y^3 = x^2y^4
yx^(-3) + xy^(-3) = x^2y^4
D(yx^(-3)) + D(xy^(-3)) = D(x^2y^4)
y’x^(-3) + y(-3x^(-4)) + 1y^(-3) + x(-3y^(-4))y’ = 2xy^4 + x^2.4y^3.y’
now move all y’ to left and simplify
y’.x^(-3) - 3xy^(-4).y’ - 4x^2y^3.y’ = 3x^(-4).y - y^(-3) + 2xy^4
y’(x^(-3) - 3xy^(-4) - 4x^2.y^3) = 3x^(-4).y - y^(-3) + 2xy^4
then divide both sides by the big ugly stuff on the left leaving just y’. I have not done this as it would be too messy in this format but I’m sure you can picture the result.
So why did I not come to the same conlclusion as the linked page above? All I did was start differentiating before simplifying which shouldn’t change the result. I must have made a mistake, but where.

PS: Wolfram alpha seems to have a 3rd answer to the problem, which doesn’t help at all.

2

The answer you’re getting is actually equivalent the linked solution for points satisfying the original equation.

In other words you can simplify your solution so there are only positive exponents and then make substitutions based on x^4 + y^4 = x^5 y^7 (equivalent to the original equation almost everywhere) to get the provided solution.

3

Yeah, the answer you’re getting will be a fractional expression that has some negative exponents appearing in the numerator and denominator, which isn’t incorrect but is messy and can be simplified. I haven’t checked it out, but I think that, when simplified, your solution should be equivalent to the given one.

4

It will require more than just simplification. It also requires substitutions based on the original equation.

5

I thought of simplifying to get there but couldn’t get it to work. Some of the exponents lined up but the co-efficients didn’t look right and there were 3 terms in the numerator and denominator.

I would be very grateful if you could show me how to do it.

6

Let’s just look at the numerator to keep the clutter down.

After ridding ourselves of negative exponents we get:

y(2 x^5 y^7 - x^4 + 3 y^4)

From the original equation we have:

y^4 = x^5 y^7 - x^4

Substituting that into the rightmost term of the first equation we get a numerator of:

y(2 x^5 y^7 - x^4 + 3(x^5 y^7 - x^4)) = y(5 x^5 y^7 - 4 x^4) = xy(5 x^4 y^7 - 4 x^3)

A similar substitution yields a denominator of :

xy(4 y^3 - 7 x^5 y^6)

Canceling xy gives the desired result.