given a density $f(x)$ of $X$ (the third moment exists), the density of $X_{\beta}$ is given by $g_{\beta}(x) = \beta f(\beta x)$, $\beta >0$. How can I show that the variation coefficient and skewness both are independent of $\beta$?
My idea was to write down first of all the expecation of $X_{\beta}$, i.e.
$$E[X_{\beta}] = \int_{-\infty}^{\infty} x g_{\beta}(x) dx = \beta \int_{-\infty}^{\infty} x f(\beta x) dx=...$$
How should I proceed now?
You can either change variable in integral $t=\beta x$ or prove that $X_\beta$ is distributed as $X/\beta$.
Indeed,
$$F_{X/\beta}(x)=P(X/\beta \leq x)=P(X\leq \beta x)= F_X(\beta x),$$ where $F_X(t)$ is CDF for $X$. Then probability density function for $X/\beta$ is equal to $$\dfrac{d}{dx}F_{X/\beta}(x)=\dfrac{d}{dx} F_X(\beta x) = \beta f(\beta x)=g_\beta(x).$$
Then for skewness one get $$ \gamma_1=\dfrac{E\left[\left(X_\beta-E[X_\beta]\right)^3\right]}{\left(Var [X_\beta]\right)^{3/2}}=\dfrac{E\left[\left(X/\beta-E[X/\beta]\right)^3\right]}{\left(Var [X/\beta]\right)^{3/2}} $$ Then use properties of expectation and variance: $$E[cX]=cE[X],\quad Var[cX] = c^2 Var[X]$$ Then reduce common factor in numerator and denominator. And $\beta$ should disappeared.