so is the simplified version of Plato's answer 7/(b/3)+4?

No, not at all, and please use grouping symbols properly. You mean 7/{(b/3) + 4}. You need to review PEMDAS.

Here is your error

$\dfrac{\dfrac{b}{3} + 7}{\left ( \dfrac{b}{3} \right )^2 + 4}$

**$\ne$** $\dfrac{\cancel {\dfrac{b}{3}} + 7}{\left ( \cancel {\dfrac{b}{3}} * \dfrac{b}{3} \right ) + 4} = \dfrac{7}{\dfrac{b}{3} + 4}.$

**THIS IS WRONG.**
For example

$\dfrac{3 + 12}{3 + 2} = \dfrac{15}{5} = 3\ NOT\ \dfrac{\cancel 3 + 12}{\cancel 3 + 2} = \dfrac{12}{2} = 6.$

You must turn both numerator and denominator into complete fractions, and then invert the denominator to multiply by the numerator.

$\dfrac{\dfrac{b}{3} + 7}{\left ( \dfrac{b}{3} \right )^2 + 4} = \dfrac{\dfrac{b}{3} + \dfrac{21}{3}}{\dfrac{b^2}{9} + 4} =$

$\dfrac{\dfrac{b + 21}{3}}{\dfrac{b^2}{9} + \dfrac{36}{9}} = \dfrac{\dfrac{b + 21}{3}}{\dfrac{b^2 + 36}{3 * 3}} =\dfrac{b + 21}{\cancel 3} * \dfrac{\cancel 3 * 3}{b^2 + 36} = \dfrac{3(b + 21)}{b^2 + 36}.$