[tex]e^{x^2y} - e^y = e^y(e^{x^2} - 1) = x[/tex]
We should ISOLATE x
[tex]e^y= \frac{x}{e^{x^2} - 1}[/tex]
Find the Natural Log of Both Sides to Make the Left Side "y"
[tex]y = ln(\frac{x}{e^{x^2}-1})[/tex]
Now, FIND THE DERIVATIVE Using Chain Rule!!!
[tex]y' = \frac{1}{\frac{x}{e^{x^2}-1}} * \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)^2}= {\frac{e^{x^2}-1}{x}}* \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)^2} = {\frac{1}{x}}* \frac{(1(e^{x^2}-1) - x(2x*e^{x^2})}{(e^{x^2}-1)} = {\frac{1}{x}}* \frac{(e^{x^2}-1 - 2x^2e^{x^2})}{(e^{x^2}-1)}[/tex]
