Calculating elasticity for $y^2e^{x+\frac{1}{y}}=3$ I want to calculate elasticity of y with respect to x  for $$y^2e^{x+\frac{1}{y}}=3$$
My attempt:
I calculate $y'$ using:
$$y'= -\frac{f_1'(x,y)}{f_2'(x,y)} = - \frac{y}{2-y}$$
I calculated this using implicit differentiation that can be used where $f(x,y)=c$, in my case $f(x,y)= y^2e^{x+\frac{1}{y}}$ and $c=3$.
Then elasticity should be just
$$y'\frac{x}{y} = -\frac{x}{2-y}$$
But I was told by professor that is a wrong answer  and that correct answer is: $\frac{xy}{1-2y}$.
I don't understand where I have made a mistake and how come its  $\frac{xy}{1-2y}$. Please can someone help me to understand it?
$$y'\frac{x}{y}= -\frac{f_1'(x,y)}{f_2'(x,y)}\frac{x}{y} = - \frac{y^2\exp(x+1/y)}{2y\exp(x+1/y)+y^2\exp(x+1/y)(-y^{-2})}\,\frac{x}{y}=\\=\frac{y^2}{1-2y}\,\frac{x}{y}=\frac{xy}{1-2y}.$$