How to sketch an $xy$ graph from a $yu$ and $xu$ graph. Question
Given the graphs of two functions, $u=f(y)$ on $yu$ axes and $u=g(x)$ on $xu$ axes, what methods are there to sketch the graph $f(y) = g(x)$ on $xy$ axes?
Furthermore, is there any intuition on what this new graph is showing?
Example
Given graphs of $u=y^{-1}e^y$ and $u=8(x^3-x)$, sketch $y^{-1}e^y = 8(x^3-x)$.
From their equations and graphs (as sketched by Desmos),
$u=8(x^3-x)$ and  $u=y^{-1}e^y$,
how can I derive the "combined" graph of $y^{-1}e^y = 8(x^3-x)$?
Thoughts and Progress
Through online graphing software, the task is trivial. I am more interested in how this could be done by hand and how the two graphs are "combined" to create the new one.
My intuition leads me to believe this work would: You could visualize $f(y)$ and $g(x)$ as planes on a set of $xyu$ axes. Then, marking out the points of intersection between the two planes (that is, the points which satisfy $f(y) = g(x)$) you could view this from the $xy$ plane "perspective" (i.e. so that the $x$ and $y$ axes are both perpendicular to your line of sight). The resulting marked points would display the graph in question.
However my ability (and I imagine most other people's ability) to visualize three-dimensional objects is not nearly good enough to make this technique viable; I am convinced there is a better method for deriving the resulting graph by hand.
I am assuming the reader is familiar with standard graph sketching techniques such as examining asymptotic behavior, turning points, and convexity.
 A: For how to think about it, consider that both $u=8(x^3-x)$ and $u=y^{-1}e^y$ can be treated as $3$-d graphs, where in the first graph $y$ takes on any value, and so looking from the $xy$ perspective we get what looks like the plane completely filled, and from the second graph we have the same for $x$, and so the plane looks full again.  The intersection of these two graphs by lining up the axes and viewing from the $xy$ perspective is the graph of $8(x^3-x)=y^{-1}e^y$.
Performing this intersection by hand is tedious, but interval arithmetic is an effective tool.  I will demonstrate just a small part of how this is done for the area given by $(x,y)\in [1,2]^2$.
First, verify that the graph passes through this area.  So for the $x$ portion we have
$$\begin{align}
&8([1,2]^3-[1,2])\\=&8([1,8]-[1,2])\\=&8([-1,7])\\=&[-8,56]
\end{align}$$
and the $y$ portion becomes
$$\begin{align}
&[1,2]^{-1}e^{[1,2]}\\
=&\left[\frac 12,1\right]\cdot[e^1,e^2]\\
=&\left[\frac e2,e^2\right]
\end{align}$$
The decision of whether the graph passes through this area then becomes a question of whether the two intervals shown above have an overlap.  Since $56 \ge e^2$ and $-8\le \frac e2$, the overlap is present.  Now we subdivide the area and check each subdivision.  This particular equation has the variables separated, so if I choose to subdivide only one direction at a time, the interval associated with the other variable does not change.  I will choose to subdivide the $x$ interval in half, as $x\in[1,1.5]$ and $x\in[1.5,2]$.  I may end up measuring a portion of the graph multiple times by including boundaries in every subdivision, but this does not hurt the result.  Taking the $x$ side of the equation again, we first have
$$8([1,1.5]^3-[1,1.5])=8([1,3.375]-[1,1.5])\\
=8([-0.5,2.375])=[-4,19]$$
which again clearly overlaps $\left[\frac e2,e^2\right]$.  The other subdivision works out as
$$8([1.5,2]^3-[1.5,2])=8([3.375,8]-[1.5,2])\\
=8([1.375,6.5])=[11,52]$$
Now, as $e^2\lt 9\lt 11$, we see that this interval does not overlap $\left[\frac e2,e^2\right]$, so we know that the graph does not pass through this portion of the area.
It should be fairly clear how easy this approach is, but the degree of tedium is significant.  This method is very good for a computer to use for graphing for these reasons, plus the fact that the rules of interval arithmetic make it so that it is possible to "prove" that a graph matches a given function.  There are no limits on how many variables and how complex the equation is for interval arithmetic to be applied, the only limitation is whether a given equation will produce usable results in a reasonable time frame.  I don't have specific examples on hand, but there were many times where I was dealing with complex functions for which my favorite graphing program "Grafeq" would fail to obtain results.
A: The system 
\begin{align}
y^{-1}\exp(y) &= 8(x^3-x) 
\end{align}
has a solution for $y$ in terms of Lambert W function,
\begin{align}
-y\exp(-y) &= -\frac{1}{8(x^3-x)} 
,\\
\operatorname{W}(-y\exp(-y))
 &= \operatorname{W}\left(-\frac{1}{8(x^3-x)}\right) 
,\\
y&=
-\operatorname{W}\left(-\frac{1}{8(x^3-x)}\right)
\tag{1}\label{1}
.
\end{align}  
For real solutions 
we need to consider two branches
\begin{align}
y_0(x)&=
-\operatorname{W_0}\left(-\frac{1}{8(x^3-x)}\right)
\tag{2}\label{2}
,\\
\text{and }\quad
y_1(x)&=
-\operatorname{W_{-1}}\left(-\frac{1}{8(x^3-x)}\right)
\tag{3}\label{3}
.
\end{align}
But first, we need to consider 
the argument $t$ of $\operatorname{W}$:
\begin{align}
t(x)&=-\frac{1}{8(x^3-x)}
=-\frac{1}{8x(x-1)(x+1)}
\tag{4}\label{4}
.
\end{align}
It is not difficult to make a sketch of \eqref{4}:

$$
\begin{array}{ccccc} 
\hline
&x-\text{range}  & t-\text{range} & y(x) & y-\text{range}
\\ \hline
1&(-\infty,-1) & (0,\infty) & \color{red}{-\operatorname{W_0}(t)} & \color{red}{(-\infty,0)}
\\ 
2&(-1,x_1) & (-\infty,-\tfrac1{\mathrm{e}}) & - & -
\\ 
3&\text[x_1,x_2\text] & \text[-\tfrac1{\mathrm{e}},-\tfrac{3\sqrt3}{16}\text] & \color{red}{-\operatorname{W_0}(t)},
\ \color{blue}{-\operatorname{W_{-1}}(t)}
& \color{red}{\text[-\operatorname{W_0}(-\tfrac{3\sqrt3}{16}),1\text]}
\cup
\color{blue}{(1,-\operatorname{W_{-1}}(-\tfrac{3\sqrt3}{16})\text]}
\\ 
4&(x_2,0) & (-\infty,-\tfrac1{\mathrm{e}}) & - & -
\\ 
5&(0,1) & (\tfrac{3\sqrt3}{16},\infty) & \color{red}{-\operatorname{W_0}(t)} &\color{red}{(-\infty,-\operatorname{W_0}(\tfrac{3\sqrt3}{16})\text]}
\\ 
6&(1,x_3) & (-\infty,-\tfrac1{\mathrm{e}}) & - & -
\\ 
7&\text[x_3,\infty) & \text[-\tfrac1{\mathrm{e}},0) 
& \color{red}{-\operatorname{W_0}(t)},
\ \color{blue}{-\operatorname{W_{-1}}(t)} 
& \color{red}{(0,1]}\cup\color{blue}{(1,\infty)}
\\ \hline
\end{array}
$$

Values of $x_1,x_2,x_3$
are solutions of 
\begin{align}
t(x)&=-\frac1{\mathrm{e}}
,
\end{align}
or the cubic roots of
\begin{align}
8x^3-8x-\mathrm{e}&=0
,
\end{align}
and can be found as
\begin{align}
x_1&=-\tfrac{2\sqrt3}3\cos\left(\tfrac\pi3
-\tfrac13\arctan\left(\sqrt{\tfrac{256}{27\mathrm{e}^2}-1}\right)\right)
\approx -0.731986
,\\
x_2&=-\tfrac{2\sqrt3}3\cos\left(\tfrac\pi3
+\tfrac13\arctan\left(\sqrt{\tfrac{256}{27\mathrm{e}^2}-1}\right)\right)
\approx -0.4074
,\\
x_3&=\tfrac{2\sqrt3}3\cos\left(
\tfrac13\arctan\left(\sqrt{\tfrac{256}{27\mathrm{e}^2}-1}\right)\right)
\approx 1.13939
.
\end{align}
