Finding Moment Generating Function of Normal Distribution I need to show that the moment generating function of $Y$ is $$M(t)=(1 − 2σ^{2}t)^{−1/2}$$ where $X$ ∼ $N$($0$, $σ^{2}$) and that $Y$ = $X^{2}$.
I know the moment generating function of $Z$ is $$E(e^{tz})= \int_\mathbb{R}\ e^{tz} f(z)dz$$
Also since $X$ is a normal distribution (with mean $0$ and standard deviation $σ$), $$f(x) = \frac{1}{\sqrt{2\piσ^{2}}}exp(\frac{-x^{2}}{{2σ^{2}}})$$
I think $M(t)=E(e^{tx^{2}})$, so $M(t)=\frac{1}{\sqrt{2\piσ^{2}}}\int_{-\infty}^{\infty} e^{x^{2}(t-\frac{1}{2σ^{2}})} dx$
But then what's next? Am I supposed to use the Guassian function? Any help would be appreciated thanks. 

Yes you need to use a Gaussian distribution. As you started:
$$
\begin{align}
M(t) = \mathbb{E}(e^{tx^{2}}) &= \frac{1}{\sqrt{2\pi}\sigma}\int_{\mathbb{R}}e^{x^{2}t}e^{-\frac{1}{2}\frac{x^{2}}{\sigma^{2}}}dx \\
&= \frac{1}{\sqrt{2\pi}\sigma}\int_{\mathbb{R}}e^{-\frac{1}{2}x^{2}\frac{1-2\sigma^{2}t}{\sigma^{2}}}dx\\
&=\frac{1}{\sqrt{2\pi}\sigma}\int_{\mathbb{R}}e^{-\frac{1}{2}\big(x\frac{\sqrt{1-2\sigma^{2}t}}{\sigma}\big)^{2}}dx
\end{align}
$$
Then you apply the following change of variable:
$$
z = x\frac{\sqrt{1-2\sigma^{2}t}}{\sigma}
$$
$$
\begin{align}
M(t) &= \frac{1}{\sqrt{1-2\sigma^{2}t}}\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-\frac{1}{2}z^{2}}dz\\
&= (1-2\sigma^{2}t)^{-{1}/{2}}
\end{align}
$$
Alternatively, you can also recognize that a squared Normal random variable follows Chi-squared ($\chi^{2}$) distribution and compute the moment generating function of this distribution with $k=1$ ($\chi^{2}$).
