The condition given is not a Robin type of condition. Robin conditions involve the value of the function and the first derivative at the same endpoint. You have a Dirichlet and a Neumann condition, as well as an initial condition.
This problem can be solved by separation of variables.
$$
\frac{T''(t)}{9T(t)} = \lambda, \;\; \lambda = \frac{X''(x)}{X(x)}
$$
with spatial conditions
$$
X'(0)=0,\;\; X(1)=0.
$$
The eigenfunctions of this problem have the form
$$
X_n(x) = \cos(\lambda_n x),\;\;\;\lambda_n = \frac{\pi}{2}+n\pi,\;\; n=0,1,2,\cdots.
$$
The corresponding solutions $T_n$ are
$$
T_n(t) = A_n\cos(3\lambda_n t)+B_n\sin(3\lambda_n t).
$$
So the general solution is
$$
u(t,x)=\sum_{n=1}^{\infty}(A_n\cos(3\lambda_n t)+B_n\sin(3\lambda_n t))\cos(\lambda_n x).
$$
There are not enough conditions to uniquely solve the problem because the one $t=0$ condition does not involve the constants $B_n$. However, one solution is obtained by setting all $B_n=0$, which is equivalent to imposing the additional condition $u_t(0,x)=0$. The remaining condition is
$$
2\cos(3\pi x/2) = \sum_{n=1}^{\infty} A_n\cos(\lambda_n x)
= \sum_{n=1}^{\infty}A_n\cos((n+1/2)\pi x).
$$
The eigenfunctions $\{ \cos((n+1/2)\pi x) \}_{n=1}^{\infty}$ are mutually orthogonal on $[0,1]$, which allows you to isolate the $A_n$ using the above condition.
