The PDF of a Normal distribution is
$$f_{\mu, \sigma}(x) = \frac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{(x-\mu )^2}{2 \sigma ^2}}dx$$
but in terms of $\tau = 1/\sigma^2$ it is
$$g_{\mu, \tau}(x) = \frac{\sqrt{\tau}}{\sqrt{2 \pi}} e^{-\frac{\tau(x-\mu )^2}{2 }}dx.$$
The PDF of a Gamma distribution is
$$h_{\alpha, \beta}(\tau) = \frac{1}{\Gamma(\alpha)}e^{-\frac{\tau}{\beta }} \tau^{-1+\alpha } \beta ^{-\alpha }d\tau.$$
Their product, slightly simplified with easy algebra, is therefore
$$f_{\mu, \alpha, \beta}(x,\tau) =\frac{1}{\beta^\alpha\Gamma(\alpha)\sqrt{2 \pi}} e^{-\tau\left(\frac{(x-\mu )^2}{2 } + \frac{1}{\beta}\right)} \tau^{-1/2+\alpha}d\tau dx.$$
Its inner part evidently has the form $\exp(-\text{constant}_1 \times \tau) \times \tau^{\text{constant}_2}d\tau$, making it a multiple of a Gamma function when integrated over the full range $\tau=0$ to $\tau=\infty$. That integral therefore is immediate (obtained by knowing the integral of a Gamma distribution is unity), giving the marginal distribution
$$f_{\mu, \alpha, \beta}(x) = \frac{\sqrt{\beta } \Gamma \left(\alpha +\frac{1}{2}\right) }{\sqrt{2\pi } \Gamma (\alpha )}\frac{1}{\left(\frac{\beta}{2} (x-\mu )^2+1\right)^{\alpha +\frac{1}{2}}}.$$
Trying to match the pattern provided for the $t$ distribution shows there is an error in the question: the PDF for the Student t distribution actually is proportional to
$$\frac{1}{\sqrt{k} s }\left(\frac{1}{1+k^{-1}\left(\frac{x-l}{s}\right)^2}\right)^{\frac{k+1}{2}}$$
(the power of $(x-l)/s$ is $2$, not $1$). Matching the terms indicates $k = 2 \alpha$, $l=\mu$, and $s = 1/\sqrt{\alpha\beta}$.
Notice that no Calculus was needed for this derivation: everything was a matter of looking up the formulas of the Normal and Gamma PDFs, carrying out some trivial algebraic manipulations involving products and powers, and matching patterns in algebraic expressions (in that order).
Best Answer
Assume $\nu \gt 2$ so that this distribution actually has a mean and standard deviation (otherwise you cannot standardize it). By direct calculation, its mean equals $\mu$ and its variance equals $\nu / (\lambda (\nu-2))$. Standardizing it, by construction, creates a distribution of the same shape but zero mean and unit standard deviation. Thus, for the standardized distribution, the formula is the same but we must have $\mu = 0$ and $\nu / (\lambda (\nu-2)) = 1$; that is, $\lambda = \nu/(\nu-2)$. Whence, plugging these values into the formula,
$$f_Z(z) = \frac{\Gamma(\frac{\nu + 1}{2})}{\Gamma(\frac{\nu}{2})} \frac{1}{\sqrt{\pi(\nu-2)}} \left[1+\frac{z^2}{\nu-2}\right]^{-\frac{\nu+1}{2}}.$$
This assumes $\mathbb{E}[X] = \mu$ and $\mathbb{E}[(X-\mu)^2]$ are known in advance, not estimated from data.