The given equation is $9 x^{2}-3 k x+k=0$.

This is of the form $a x^{2}+b x+c=0$, where $a=9, b=-3 k$ and $c=k$.

$\therefore D=b^{2}-4 a c=(-3 k)^{2}-4 \times 9 \times k=9 k^{2}-36 k$

The given equation will have real and equal roots if $D=0$.

$\begin{array}{l}
\therefore 9 k^{2}-36 k=0 \\
\Rightarrow 9 k(k-4)=0 \\
\Rightarrow k=0 \text { or } k-4=0 \\
\Rightarrow k=0 \text { or } k=4
\end{array}$

But, $k \neq 0$ (Given)

Hence, the required values of $k$ is 4 .