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Prove that $\forall x\in \mathbb{R}$ there is only one $y=h(x)$ so that $$y+x^2y^3+x+y^5x^4=1$$

I am currenty studying multi-variable calculus on my own and I have no clue how to solve this problem. Which results of multi-variable calculus do I have to study to solve it? I would appreciate if someone could solve it or share similiar problems to this one.

codingnight
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$\,x = 0 \implies y=1\,$ i.e. $\,h(0)=1\,$. For $\,x \ne 0\,$, let $\,z=xy\,$, then multiply by $\,x\,$, and write it as:

$$ z^5+z^3+z = x - x^2 $$

The LHS $\,f(z)=z^5+z^3+z\,$ is continuous and strictly increasing on $\,\mathbb{R}\,$, and since the limits at $\,\pm \infty\,$ are $\,\pm\infty\,$ respectively, it follows that $\,f : \mathbb{R} \to \mathbb{R}\,$ is a bijection. Then the unique solution of the equation is $\displaystyle\,z = f^{-1}(x-x^2) \iff y = \frac{f^{-1}(x-x^2)}{x} = h(x)\,$.

[ EDIT ]   Following up on the comments, such problems would be solved in the general case using the implicit function theorem. Here in particular $\,f(x,y) = y^5x^4+y^3x^2+y+x-1\,$, and $\displaystyle\,\frac{\partial{f}}{\partial{y}}(x,y)=5y^4x^4+3y^2x^2+1 \ge 1\,$ is never $\,0\,$, so the map $\,y=h(x)\,$ is defined at all points.
dxiv
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  • Wow. This is an amazing answer, but I suspect like the OP I would never have come to an answer like this. If you have time, can you give some insight about what you noticed apriori that allowed you to devise this solution? – IntegrateThis Jan 06 '18 at 04:20
  • Yes please, what should I be looking for when solving this problems? – codingnight Jan 06 '18 at 04:31
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    @codingnight In the general case, you would look at the implicit function theorem. I'll add a note to the answer about that. However, given the particular form of the equation here, my guess is that it was meant to be solved using a shortcut such as the one above, rather than the full weight of the implicit function theorem. The clue for me was that all terms in $,y,$ contained a power of $,x,$ exactly one less. – dxiv Jan 06 '18 at 04:38
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    @IntegrateThis See my previous comment, and the edit added at the end. – dxiv Jan 06 '18 at 04:51