The original sum can be written as
$$T(\alpha,\beta,\gamma,n)=\sum_{x,y\le n}x^\alpha y^\beta(x,y)^\gamma,$$
where $(x,y)=\mathrm{gcd}(x,y)$. One can find asymptotic formula for this sum using standart approach. Let $d=(x,y)$. Then
$$T(\alpha,\beta,\gamma,n)=\sum_{d\le n}d^\gamma\sum_{{x,y\le n\atop (x,y)=d}}x^\alpha y^\beta=\sum_{d\le n}d^{\alpha+\beta+\gamma}\sum_{{x,y\le n/d\atop (x,y)=1}}x^\alpha y^\beta.$$
The condition $(x,y)=1$ can be removed using Möbius function:
$$T(\alpha,\beta,\gamma,n)=\sum_{d\le n}d^{\alpha+\beta+\gamma}\sum_{\delta\le n/d}\mu(\delta)\sum_{{x,y\le n/d\atop \delta\mid(x,y)}}x^\alpha y^\beta=\sum_{d\le n}d^{\alpha+\beta+\gamma}\sum_{\delta\le n/d}\mu(\delta)\delta^{\alpha+\beta}\sum_{x,y\le n/(d\delta)}x^\alpha y^\beta.$$
The last sum (for $\alpha,\beta>-1$) is $\sim\frac{n^{\alpha+\beta+2}}{(\alpha+1)(\beta+1)(d\delta)^{\alpha+\beta+2}},$ so (for $\gamma>1$)
$$T(\alpha,\beta,\gamma,n)\sim \frac{n^{\alpha+\beta+\gamma+1}}{\zeta(2)(\alpha+1)(\beta+1)}.$$