Show that $e^{xy}+y=x-1$ is an implicit solution to the differential equation $\frac{dy}{dx} = \frac{e^{-xy}-y}{e^{-xy}+x}$

Question

I began by using implicit differentiation on $e^{xy}+y=x-1$.

From that I got:

$$\left(y+x\frac{dy}{dx}\right)e^{xy}+\frac{dy}{dx}=1$$

Then using algebra I got to the point where I had this equation:

$$\frac{dy}{dx}=\frac{1-ye^{xy}}{xe^{xy}+1}$$

I'm not sure if I messed up somewhere along the road, or if my final equation is actually the equivalent to $\dfrac{dy}{dx} = \dfrac{e^{-xy}-y}{e^{-xy}+x}$ but I would like help knowing either where I went wrong or how to convert my equation to the final answer.

Answer 1

we have $$e^{xy}y+e^{xy}y'x+y'=1$$so we get $$y'(e^{xy}x+1)=1-ye^{x}$$ $$y'=\frac{1-ye^{xy}}{1+x^{xy}}$$ multiplying numerator and denominator by $$e^{-xy}$$ we get $$y'=\frac{e^{-xy}-y}{e^{-xy}+x}$$ this is what we want to prove