No, because the k in the numerator and the k in the denominator are the same k. They would both have to change.If you change the K in the denominator to +K, then the answer is y = -2...

I agree with you up to here:

$\displaystyle \displaystyle -2\,\frac{e^{-\frac{1}{2}x^2} + K}{e^{-\frac{1}{2}x^2} - K}.$

Replacing $\displaystyle K$ with $\displaystyle -K$ yields

$\displaystyle \displaystyle -2\,\frac{e^{-\frac{1}{2}x^2} - K}{e^{-\frac{1}{2}x^2} + K}.$

Multiplying the negative sign in yields

$\displaystyle \displaystyle 2\,\frac{K-e^{-\frac{1}{2}x^2}}{K+e^{-\frac{1}{2}x^2}}.$

Now, finally, one more trick: replacing $\displaystyle K$ with $\displaystyle 1/K,$ will, I think, finish off the problem.