Topic.
I took a stab at it, and my guess is that you use the chain rule, but it looks messed up in the end.
Any help would be greatly appreciated.
I assume you mean $\displaystyle e^{xy}$ rather than $\displaystyle ye^x$
Let $\displaystyle f(x) = e^{xy}$ and then taking logs of both sides (in base e) gives:
$\displaystyle \ln{f(x)} = xy$
Now we differentiate implicitly and with the product rule recalling that the differentiating $\displaystyle \ln{f(x)} = \frac{f\prime(x)}{f(x)}$:
Right hand side: u=x and v=y so du/dx = 1 and dv/dx = dy/dx which gives us the right hand side differential as $\displaystyle y + x\frac{dy}{dx}$
$\displaystyle \frac{f\prime(x)}{f(x)} = y + x\frac{dy}{dx} $
As we know f(x) = e^(xy) we can say that $\displaystyle \frac{f\prime(x)} = ye^{xy} + {xe^{xy}\frac{dy}{dx}$
We can now collect dy/dx terms (and in this case $\displaystyle f\prime{x}$ can also be treated as dy/dx)
$\displaystyle \frac{dy}{dx} - {xe^{xy}}\frac{dy}{dx} = ye^(xy)$
Factor on the left:
$\displaystyle \frac{dy}{dx}(1 - {xe^{xy}}) = ye^(xy)$
and finally divide by $\displaystyle (1 - {xe^{xy}})$
$\displaystyle \frac{dy}{dx} = \frac{ye^(xy)}{(1 - {xe^{xy}})}$