Find the partial fraction decomposition.
\[
\frac{9 x}{x^{2}+x-2}
\]
\[
\frac{9 x}{x^{2}+x-2}=
\]

Step 1 ：Factor the denominator of the given fraction. The denominator \(x^{2}+x-2\) can be factored as \((x-1)(x+2)\).

Step 2 ：Rewrite the given fraction as \[\frac{9 x}{(x-1)(x+2)}\].

Step 3 ：Express the fraction as the sum of two simpler fractions (the partial fractions), where the denominators are the factors of the original denominator: \[\frac{9 x}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}\].

Step 4 ：Multiply through by the common denominator \((x-1)(x+2)\) to get rid of the fractions: \[9x = A(x+2) + B(x-1)\].

Step 5 ：Solve for A and B by substituting suitable values for x. If we let x = 1, the equation becomes \[9(1) = A(1+2) + B(1-1)\]. Solving this gives \(A = 3\).

Step 6 ：If we let x = -2, the equation becomes \[9(-2) = A(-2+2) + B(-2-1)\]. Solving this gives \(B = -6\).

Step 7 ：The partial fraction decomposition of the given fraction is \[\frac{9 x}{x^{2}+x-2} = \frac{3}{x-1} - \frac{6}{x+2}\].

Step 8 ：Check the work by adding the two fractions on the right side and simplifying. The result should be the original fraction. \[\frac{3}{x-1} - \frac{6}{x+2} = \frac{3(x+2) - 6(x-1)}{(x-1)(x+2)} = \frac{9x}{x^{2}+x-2}\].

Step 9 ：So, the solution is correct. The final answer is \[\boxed{\frac{9 x}{x^{2}+x-2} = \frac{3}{x-1} - \frac{6}{x+2}}\].