To solve the system of equations:

1. \(\frac{4}{3} \left( \frac{1}{x} + \frac{1}{y} \right) = 1\)
2. \(\frac{1}{6x} + \frac{1}{5y} = \frac{2}{15}\)

Let's start by simplifying and solving these equations step by step.

### Equation 1:
\[
\frac{4}{3} \left( \frac{1}{x} + \frac{1}{y} \right) = 1
\]

Multiply both sides by \(\frac{3}{4}\) to isolate the term inside the parentheses:
\[
\frac{1}{x} + \frac{1}{y} = \frac{3}{4}
\]

### Equation 2:
\[
\frac{1}{6x} + \frac{1}{5y} = \frac{2}{15}
\]

Let's rewrite this equation in terms of \(\frac{1}{x}\) and \(\frac{1}{y}\). Notice that:
\[
\frac{1}{6x} = \frac{1}{6} \cdot \frac{1}{x} \quad \text{and} \quad \frac{1}{5y} = \frac{1}{5} \cdot \frac{1}{y}
\]

So, the equation becomes:
\[
\frac{1}{6} \cdot \frac{1}{x} + \frac{1}{5} \cdot \frac{1}{y} = \frac{2}{15}
\]

Let \(a = \frac{1}{x}\) and \(b = \frac{1}{y}\). Substituting these into the equations, we get:

1. \(\frac{4}{3}(a + b) = 1\)
2. \(\frac{a}{6} + \frac{b}{5} = \frac{2}{15}\)

### Simplify Equation 1:
\[
\frac{4}{3}(a + b) = 1 \implies a + b = \frac{3}{4}
\]

### Simplify Equation 2:
\[
\frac{a}{6} + \frac{b}{5} = \frac{2}{15}
\]

To eliminate the fractions, multiply through by 30 (the least common multiple of 6 and 5):
\[
30 \left( \frac{a}{6} + \frac{b}{5} \right) = 30 \cdot \frac{2}{15}
\]

This simplifies to:
\[
5a + 6b = 4
\]

Now we have the system of linear equations:
1. \(a + b = \frac{3}{4}\)
2. \(5a + 6b = 4\)

### Solve the System:
First, solve the first equation for \(a\):
\[
a = \frac{3}{4} - b
\]

Substitute \(a\) into the second equation:
\[
5 \left( \frac{3}{4} - b \right) + 6b = 4
\]

Distribute and simplify:
\[
\frac{15}{4} - 5b + 6b = 4
\]

Combine like terms:
\[
\frac{15}{4} + b = 4
\]

Subtract \(\frac{15}{4}\) from both sides:
\[
b = 4 - \frac{15}{4} = \frac{16}{4} - \frac{15}{4} = \frac{1}{4}
\]

Now substitute \(b = \frac{1}{4}\) back into \(a = \frac{3}{4} - b\):
\[
a = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}
\]

So, \(a = \frac{1}{2}\) and \(b = \frac{1}{4}\).

Recall that \(a = \frac{1}{x}\) and \(b = \frac{1}{y}\):
\[
\frac{1}{x} = \frac{1}{2} \implies x = 2
\]
\[
\frac{1}{y} = \frac{1}{4} \implies y = 4
\]

### Solution:
The solution to the system of equations is:
\[
x = 2, \quad y = 4
\]