To solve the equations you've provided, let's break them down step by step.

Starting with the first equation:

\[
x^2 + \frac{1}{9} = \frac{5}{3}
\]

1. Subtract \(\frac{1}{9}\) from both sides:

\[
x^2 = \frac{5}{3} - \frac{1}{9}
\]

2. To subtract these fractions, we need a common denominator. The common denominator of 3 and 9 is 9. Thus, we convert \(\frac{5}{3}\):

\[
\frac{5}{3} = \frac{15}{9}
\]

Now we can perform the subtraction:

\[
x^2 = \frac{15}{9} - \frac{1}{9} = \frac{14}{9}
\]

3. Next, take the square root of both sides:

\[
x = \pm \sqrt{\frac{14}{9}} = \pm \frac{\sqrt{14}}{3}
\]

So, the solutions for the first equation are:

\[
x = \frac{\sqrt{14}}{3} \quad \text{and} \quad x = -\frac{\sqrt{14}}{3}
\]

---

Now moving on to the second equation:

\[
3x^2 + \frac{1}{9} = \frac{5}{3}
\]

1. Subtract \(\frac{1}{9}\) from both sides:

\[
3x^2 = \frac{5}{3} - \frac{1}{9}
\]

2. Again, we convert \(\frac{5}{3}\) to have a common denominator of 9:

\[
\frac{5}{3} = \frac{15}{9}
\]

Now perform the subtraction:

\[
3x^2 = \frac{15}{9} - \frac{1}{9} = \frac{14}{9}
\]

3. Divide both sides by 3:

\[
x^2 = \frac{14}{9} \cdot \frac{1}{3} = \frac{14}{27}
\]

4. Take the square root of both sides:

\[
x = \pm \sqrt{\frac{14}{27}} = \pm \frac{\sqrt{14}}{3\sqrt{3}} = \pm \frac{\sqrt{14}}{3\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \pm \frac{\sqrt{42}}{9}
\]

So the solutions for the second equation are:

\[
x = \frac{\sqrt{42}}{9} \quad \text{and} \quad x = -\frac{\sqrt{42}}{9}
\]

---

In summary, here are the solutions:

1. For \(x^2 + \frac{1}{9} = \frac{5}{3}\):
- \(x = \frac{\sqrt{14}}{3}\) and \(x = -\frac{\sqrt{14}}{3}\)

2. For \(3x^2 + \frac{1}{9} = \frac{5}{3}\):
- \(x = \frac{\sqrt{42}}{9}\) and \(x = -\frac{\sqrt{42}}{9}\)