It looks like you're trying to solve an equation that involves fractions with variables. Let's break down and solve the following equation step by step:

You wrote:

\[
\frac{X + 3}{X - 3} = \frac{3}{X^2 - 3X} + \frac{1}{X}
\]

Let's start simplifying the right side:

1. Find a common denominator for the terms on the right side. The common denominator for \(X^2 - 3X\) and \(X\) can be \(X(X - 3)\).

2. Rewrite the fractions with the common denominator:

\[
\frac{3}{X^2 - 3X} = \frac{3}{X(X - 3)}
\]

The second term becomes:
\[
\frac{1}{X} = \frac{(X - 3)}{X(X - 3)}
\]

3. Thus, combining these under the common denominator:

\[
\frac{3 + (X - 3)}{X(X - 3)} = \frac{X}{X(X - 3)}
\]

So the equation now looks like:

\[
\frac{X + 3}{X - 3} = \frac{X}{X(X - 3)}
\]

Next, we can cross-multiply:

\[
(X + 3) \cdot X = X \cdot (X - 3)
\]

Expanding both sides gives:

\[
X^2 + 3X = X^2 - 3X
\]

Now we can simplify:

1. Subtract \(X^2\) from both sides:

\[
3X = -3X
\]

2. Adding \(3X\) to both sides:

\[
3X + 3X = 0
\]

\[
6X = 0
\]

3. Dividing both sides by 6:

\[
X = 0
\]

Lastly, we should check whether this solution is acceptable (it doesn't make any denominator zero in the original equation):

Substituting \(X = 0\) back into the original equation:

\[
\frac{0 + 3}{0 - 3} = \frac{3}{0^2 - 3 \cdot 0} + \frac{1}{0}
\]

The left side is:

\[
\frac{3}{-3} = -1
\]

However, the right side becomes undefined because of the term \(\frac{1}{0}\). Thus, this solution is not valid.

### Conclusion:
The equation has no valid solution due to the division by zero.