f(x) is the pdf of two parameter gamma distribution. Gamma(alpha, beta).
$$f(x)=\frac{x^{\beta-1}*exp(-x/\alpha)}{(\alpha^\beta) *\Gamma(\beta)}$$
x>0
I need to transform x as
y=(x-mu)/sigma, x=mu+y*sigma, dx/dy=sigma
f(y)=[sigma*((mu+y*sigma)^(beta-1))exp{-(mu+ysigma)/alpha}]/{(alpha^beta) *gamma function(beta)}
Your density looks the density of a $\text{Gamma}(\beta;\alpha)$
If you want a density of a $\text{Gamma}(\alpha;\beta)$ this is the following
$$f(x)=\frac{x^{\alpha-1}\cdot \text{exp}(-x/\beta)}{\beta^\alpha\cdot \Gamma(\alpha)}$$
with
$$E(X)=\alpha\cdot\beta$$
$$V(X)=\alpha\beta^2$$
Thus using your transformation function
$$Y=\frac{X-\alpha\beta}{\beta\sqrt{\alpha}}$$
where
$$x=\beta\sqrt{\alpha}y+\alpha\beta$$
$$x'=\beta\sqrt{\alpha}$$
and thus
$$f_Y(y)=\frac{\beta\sqrt{\alpha}}{\Gamma(\alpha)\beta^\alpha}(\beta\sqrt{\alpha}y+\alpha\beta)^{\alpha-1}\text{exp}\left\{-\frac{\beta\sqrt{\alpha}y+\alpha\beta}{\beta}\right\}$$
simplifying...
$$f_Y(y)=\frac{(\sqrt{\alpha})^\alpha}{\Gamma(\alpha)}(y+\sqrt{\alpha})^{\alpha-1}\text{exp}\left\{-\sqrt{\alpha}(y+\sqrt{\alpha})\right\}\cdot\mathbb{1}_{[-\sqrt{\alpha};+\infty)}(y)$$
