Answer
$$\frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{{{\left( {{x^2} + 3} \right)}^2}\left( {5{x^2} - 3} \right)}}{{{x^2}}}$$
Work Step by Step
$$\eqalign{
& C\left( x \right) = {\left( {{x^2} + 3} \right)^3} \cr
& {\text{The average cost is defined by }}\overline C \left( x \right) = \frac{{C\left( x \right)}}{x}.{\text{ then}} \cr
& \overline C \left( x \right) = \frac{{{{\left( {{x^2} + 3} \right)}^3}}}{x} \cr
& {\text{The marginal average cost is the derivative of the average cost function}}{\text{. Then}} \cr
& {\text{the marginal average cost is given by}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{d}{{dx}}\left[ {\frac{{{{\left( {{x^2} + 3} \right)}^3}}}{x}} \right] \cr
& {\text{by using the quotient rule}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{x\frac{d}{{dx}}\left[ {{{\left( {{x^2} + 3} \right)}^3}} \right] - {{\left( {{x^2} + 3} \right)}^3}\frac{d}{{dx}}\left[ x \right]}}{{{x^2}}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{x\left( 3 \right){{\left( {{x^2} + 3} \right)}^2}\left( {2x} \right) - {{\left( {{x^2} + 3} \right)}^3}\left( 1 \right)}}{{{x^2}}} \cr
& {\text{simplify}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{6{x^2}{{\left( {{x^2} + 3} \right)}^2} - {{\left( {{x^2} + 3} \right)}^3}}}{{{x^2}}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{{{\left( {{x^2} + 3} \right)}^2}\left( {6{x^2} - {x^2} - 3} \right)}}{{{x^2}}} \cr
& \frac{d}{{dx}}\left[ {\overline C \left( x \right)} \right] = \frac{{{{\left( {{x^2} + 3} \right)}^2}\left( {5{x^2} - 3} \right)}}{{{x^2}}} \cr} $$
