Let $X\sim N(0,1)$ that is $X$ is a random variable with normal distribution with mean$=0$ and standard deviation$=1$ and $$f_X(x)=\dfrac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$$ Let $y=g(X)=\dfrac{1}{x}$. Now find the pdf for $Y$.
The first thing is that $y\in(-\infty,0)\cup(0,\infty)$ and that $y'=-1x^{-2}<0$ which means that $y$ is a decreasing function. Next $g^{-1}(y)=\dfrac{1}{y}\implies \dfrac{d}{dy}g^{-1}(y)=-1y^{-2}=\dfrac{-1}{y^2}$ But now if $x\in(0,\infty)$ then $y\in(0,\infty)$. It follows that $$f_y(y)=\dfrac{1}{\sqrt{2\pi}}e^{\dfrac{-1}{2y^2}}\left|\frac{-1}{y^2}\right|=\dfrac{1}{\sqrt{2\pi}}e^{\dfrac{-1}{2y^2}}*\frac{1}{y^2}$$ Similarly if $x\in(-\infty,0)$ then $y\in(-\infty,0)$. You get the same thing as above. Thus it follows that $$f_y(y)= \dfrac{1}{\sqrt{2\pi}}e^{\dfrac{-1}{2y^2}}\left|\frac{-1}{y^2}\right|=\dfrac{1}{\sqrt{2\pi}}e^{\dfrac{-1}{2y^2}}*\frac{1}{y^2}$$ for $y\in(-\infty,0)\cup(0,\infty)$. Would this be correct?
$\checkmark$ Yes, that looks long winded but correct.
The support for $Y$ must be $(-\infty, 0)\cup(0,\infty)$. Or equivalently: $(-\infty,\infty)\setminus\{0\}$
The transformation is:
$$\begin{align} f_Y(y) & = f_X\circ g^{-1}(y)\cdot \bigg|\frac{\mathrm d g^{-1}(y)}{\mathrm d y}\bigg| & : f_X(x) = \frac{e^{-x^2/2}}{\sqrt{2\pi}}\operatorname {\bf 1}_{(-\infty,\infty)}(x)\,,\, g(x) = x^{-1} \\[2ex] & = \frac{e^{-1/2y^2} }{ y^2\sqrt{2\pi}} \operatorname {\bf 1}_{(-\infty,0)\cup(0,\infty)}(y) & \because g^{-1}(y)=y^{-1}\,,\, (g^{-1})'(y) = -y^{-2} \end{align}$$