Given,

$\rm 3^x + 3^{x+2} = \frac{10}{3}$

By using the law of indices, $\rm a^{m+n} = a^m \cdot a^n$, we get,

$\rm or, 3^x + 3^x \cdot 3^2 = \frac{10}{3}$

$\rm or, 3^x \left ( 1 + 3^2 \right ) = \frac{10}{3}$

$\rm or, 3^x \left ( 1 + 9 \right ) = \frac{10}{3}$

$\rm or, 3^x \cdot 10 = \frac{10}{3}$

Dividing both sides by 10, we get,

$\rm or, 3^x \cdot \frac{10}{10} = \frac{10}{3} \cdot \frac{1}{10}$

$\rm or, 3^x = \frac{1}{3}$

By using the law of indices, $\rm a^{-m} = \frac{1}{a^m}$, we get,

$\rm or, 3^x = 3^{-1}$

The base of the terms in the left-hand and the right-hand sides of the equation are same, so we equate their exponents, we get,

$\rm \therefore x = -1$

Hence, the required value of x = -1.