Does any transformation of a random variable result in a valid PDF?

Question

I have a random variable $X$ with PDF $f_X(x)=\frac{e^{-x}}{(1+e^{-x})^2}, \ x\in(-\infty,\infty)$. When I apply the transformation $Y=X^2$, I obtain Y with PDF $f_Y(y)=\frac{1}{2\sqrt{y}}\left(\frac{e^{\sqrt{y}}}{(1+e^{\sqrt{y}})^2}+\frac{e^{-\sqrt{y}}}{(1+e^{-\sqrt{y}})^2}\right) \ y\in[0,\infty)$. I checked that this integrates to one, however what strikes me as strange is that the PDF is undefined at zero. What does this mean? Is this a valid PDF? Do all transformations result in valid PDFs given that the starting PDF was valid?