Applying KVL, we get

\begin{aligned} -4+\frac{V_1}{\frac{1}{2}}+\frac{V_1-V_2}{\frac{1}{3}}+\frac{V_2-V_1}{\frac{1}{3}}+\frac{V_2}{\frac{1}{6}}-9 & =0 \\ -4+2 V_1+6 V_2-9 & =0 \end{aligned}

2 V_1+6 V_2=13          (i)

Since the current i through an ideal voltage source can be of any value; it is not possible to write the nodal equations at nodes (1) and (2) independently.
Hence, a supernode procedure is followed.
Inside the supernode, applying KVL, we get

V_1-5-V_2=0

V_1-V_2=5            (ii)

Solving Eqs. (i) and (ii), we get

V_1=5.375  V \text { and } V_2=0.375  V