Like Diophantine equation The  equation $x^n - ny^x-nxy$ = $0$ has solution set $(n, x, y)  = (1, 1, \frac12), (2, 1, \frac14), (3, 1, \frac16), \ldots$
I would like to know/learn the following (Kindly discuss)
1)  If we want to know the graph. How would be the look of the graph and what kind of graph we get?
2)  The cited above equations has infinite solutions with $x = 1$. Can we have solutions with $x >1$ and other $n, y$ are some positives?
3)  If solutions exists how to find them for $x < 1$ and $x > 1$? 
Thanks in advance.

Let's take $x=2$ and see what happens. 
The equation becomes $ny^2+2ny-2^n=0$. That's a quadratic equation for $y$, so we get $$y={-2n\pm\sqrt{4n^2+(4n)2^n}\over2n}={-n\pm\sqrt{n^2+2^nn}\over n}$$ Wanting $n$ and $y$ to be positive, we reject the negative root, and get $$y={\sqrt{n^2+2^nn}-n\over n}$$ so there is a $y$ for every $n$, hence, an uncountable infinity of solutions for $x=2$. 
$x=1/2$ also leads to a quadratic in $y$ --- details left to the reader. 
If by "the graph" we mean the set of all points $(n,x,y)$ in ${\bf R}^3$ satisfying the equation, I doubt it looks like much of anything. 
