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Your input: perform the partial fraction decomposition of $$$\frac{1}{9 u^{2} - 16}$$$

Factor the denominator: $$$\frac{1}{9 u^{2} - 16}=\frac{1}{\left(3 u - 4\right) \left(3 u + 4\right)}$$$

The form of the partial fraction decomposition is

$$\frac{1}{\left(3 u - 4\right) \left(3 u + 4\right)}=\frac{A}{3 u - 4}+\frac{B}{3 u + 4}$$

Write the right-hand side as a single fraction:

$$\frac{1}{\left(3 u - 4\right) \left(3 u + 4\right)}=\frac{\left(3 u - 4\right) B + \left(3 u + 4\right) A}{\left(3 u - 4\right) \left(3 u + 4\right)}$$

The denominators are equal, so we require the equality of the numerators:

$$1=\left(3 u - 4\right) B + \left(3 u + 4\right) A$$

Expand the right-hand side:

$$1=3 u A + 3 u B + 4 A - 4 B$$

Collect up the like terms:

$$1=u \left(3 A + 3 B\right) + 4 A - 4 B$$

The coefficients near the like terms should be equal, so the following system is obtained:

$$\begin{cases} 3 A + 3 B = 0\\4 A - 4 B = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=\frac{1}{8}$$$, $$$B=- \frac{1}{8}$$$

Therefore,

$$\frac{1}{\left(3 u - 4\right) \left(3 u + 4\right)}=\frac{\frac{1}{8}}{3 u - 4}+\frac{- \frac{1}{8}}{3 u + 4}$$

Answer: $$$\frac{1}{9 u^{2} - 16}=\frac{\frac{1}{8}}{3 u - 4}+\frac{- \frac{1}{8}}{3 u + 4}$$$