Method 1. From first principles.
The adjoint is completely determined by its value on $(1,0)$ and $(0,1)$ (as is any good linear transformation with domain $\mathbb{R}^2$. So let's look at it:
$$\langle T(x,y),(1,0)\rangle = \langle (2x-y,x+3y),(1,0)\rangle = 2x-y$$
so we need
$$\langle (x,y),T^*(1,0)\rangle = 2x-y.$$
This means that we need $T(1,0) = (2,-1)$.
Similarly, since
$$\langle T(x,y),(0,1)\rangle = \langle (2x-y,x+3y),(0,1)\rangle = x+3y$$
so what should $T^*(0,1)$ be?
Method 2. From the matrix for $T$.
The matrix of $T$ with respect to the standard basis is
$$\left(\begin{array}{rr}
2& -1\\
1&3
\end{array}\right).$$
Since $[T^*] = [T]^*$, where the left hand side indicates the adjoint and the right hand side indicates the conjugate transpose matrix, then the matrix of $T^*$ is... (And of course, from the matrix we can figure out the formula.)