$e^{-xy} + e^{xy} = 2e^{-y}$ - where am I going wrong? I am trying to see if there is any $x$ (real or complex) for which this equation can be solved.
$$e^{-xy} + e^{xy} = 2e^{-y}$$
Step 1. Multiplying both sides by y,
$$ye^{-xy} + ye^{xy} = 2ye^{-y}$$
Step2. Partially differentiating the original equation (i.e. $e^{-xy} + e^{xy} = 2e^{-y}$ ) w.r.t. x,
$$-ye^{-xy} + ye^{xy} = 0$$
Step3. Adding Steps 1 and 2,
$$2ye^{xy} = 2ye^{-y}$$
this gives $x=-1$.
Obviously this is not a solution. Where am I going wrong? or is there a complex value of x for which Step 3 holds?

The other answers, but your approach in Step 2 has an error for another reason:
Even if you were looking for an $x$ such that $f(x,y)=g(x,y)$ for all $y$, this would hold if you differentiate both sides of this by $x$ and not by $y$.
