Use the binomial expansion to expand \(\left(\frac{x}{3} + \frac{2}{y}\right)^4\).

To expand \(\left(\frac{x}{3} + \frac{2}{y}\right)^4\) using the binomial expansion, we can use the formula:

\((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

In this case, \(a = \frac{x}{3}\), \(b = \frac{2}{y}\), and \(n = 4\).

So, the expansion will be:

\(\left(\frac{x}{3} + \frac{2}{y}\right)^4 = \binom{4}{0} \left(\frac{x}{3}\right)^4 \left(\frac{2}{y}\right)^0 + \binom{4}{1} \left(\frac{x}{3}\right)^3 \left(\frac{2}{y}\right)^1 + \binom{4}{2} \left(\frac{x}{3}\right)^2 \left(\frac{2}{y}\right)^2 + \binom{4}{3} \left(\frac{x}{3}\right)^1 \left(\frac{2}{y}\right)^3 + \binom{4}{4} \left(\frac{x}{3}\right)^0 \left(\frac{2}{y}\right)^4\)

Now, we can calculate each term:

\(\binom{4}{0} \left(\frac{x}{3}\right)^4 \left(\frac{2}{y}\right)^0 = 1 \cdot \frac{x^4}{81}\)

\(\binom{4}{1} \left(\frac{x}{3}\right)^3 \left(\frac{2}{y}\right)^1 = 4 \cdot \frac{x^3}{27y}\)

\(\binom{4}{2} \left(\frac{x}{3}\right)^2 \left(\frac{2}{y}\right)^2 = 6 \cdot \frac{x^2}{9y^2}\)

\(\binom{4}{3} \left(\frac{x}{3}\right)^1 \left(\frac{2}{y}\right)^3 = 4 \cdot \frac{x}{3y^3}\)

\(\binom{4}{4} \left(\frac{x}{3}\right)^0 \left(\frac{2}{y}\right)^4 = 1 \cdot \frac{16}{y^4}\)

Putting it all together, we get:

\(\frac{x^4}{81} + \frac{4x^3}{27y} + \frac{6x^2}{9y^2} + \frac{4x}{3y^3} + \frac{16}{y^4}\)

Using the binomial expansion, we have:

\begin{align-}
\left(\frac{x}{3} \frac{2}{y}\right)^4 &= \left(\frac{x}{3} \right)^4 \left(\frac{2}{y}\right)^4 \\
&= \left( \frac{x^4}{81} \right) \left( \frac{16}{y^4} \right) \\
&= \frac{16x^4}{81y^4}
\end{align-}

Therefore, \(\left(\frac{x}{3} \frac{2}{y}\right)^4 = \frac{16x^4}{81y^4}\).