A fraction problem a = x + 1 x b = y + 1 y...

Note that $c$ is not uniquely specified by $a$ and $b$, since solving $a=x+\frac{1}{x}$ for $x$ yields two solutions which are reciprocals of each other, and applying $x\mapsto \frac{1}{x}$ or $y\mapsto \frac{1}{y}$ to $xy+\frac{1}{xy}$ yields $\frac{x}{y}+\frac{y}{x}$. However, applying either substitution again gives back $xy+\frac{1}{xy}$, so $c$ can take on either of those values.
Let $c_1=xy+\frac{1}{xy}$ and $c_2=\frac{x}{y}+\frac{y}{x}$. One can check that
$c_1+c_2=ab$
and
$c_1{c}_2=a^2+b^2-4.$
Use the quadratic formula to solve for the possible values of $c$

Multiply all things by the LCD to remove fractions.
$ax=x^2+1$
$by=y^2+1$
$cxy=x^2{y}^2+1$
Multiply the first two equations to get
$abxy=x^2{y}^2+1+x^2+y^2=c+x^2+y^2$
Adding the first two equations, we get
$ax+by=x^2+y^2+2$
Combining these two lines,
$abxy=c+ax+by-2$
At this point, I'd just go back and solve for $x,y$ using the quadratic formula,
$c+a\left(\frac{a\pm \sqrt{a^2-4}}{2}\right)+b\left(\frac{b\pm \sqrt{b^2-4}}{2}\right)-2=ab\left(\frac{a\pm \sqrt{a^2-4}}{2}\right)\left(\frac{b\pm \sqrt{b^2-4}}{2}\right)$
Not the most beautiful thing in the world, but at least you don't have to square the thing or anything messy like that.