Derivative of implicit functions

Question

This problem got me confused. I'm supposed to find the value of the derivative of $$y\mathrm{e}^y=\mathrm{e}^{x+1}$$ at the point $x=0$.

I did find the derivate but I can't find its value because the derivate is $$\frac{y}{y+1}.$$ So basically there is no $x$ to plug in.

Answer 1

we get $$y'e^{y}+ye^{y}y'=e^{x+1}$$ so $$y'=\frac{e^{x+1}}{e^y(y+1)}$$ for $$y+1\neq 0$$

Answer 2

You are ok. The left side of your equation is an increasing function of y and if x=0 the right side is 1 and the left side is 1 only at y=1. $\frac{y}{y+1}$ =$\frac{1}{2}$