Answer
$$y' = {x^{5/2}}\tanh \sqrt x {\operatorname{sech} ^2}\sqrt x + 3{x^2}{\tanh ^2}\sqrt x $$
Work Step by Step
$$\eqalign{
& y = {x^3}{\tanh ^2}\left( {\sqrt x } \right) \cr
& {\text{find the derivatvive}} \cr
& y' = \left( {{x^3}{{\tanh }^2}\left( {\sqrt x } \right)} \right)' \cr
& {\text{product rule}} \cr
& y' = {x^3}\left( {{{\tanh }^2}\left( {\sqrt x } \right)} \right)' + {\tanh ^2}\left( {\sqrt x } \right)\left( {{x^3}} \right)' \cr
& {\text{chain rule}} \cr
& y' = {x^3}\left( {2\tanh \sqrt x } \right)\left( {\tanh \sqrt x } \right)' + {\tanh ^2}\left( {\sqrt x } \right)\left( {{x^3}} \right)' \cr
& {\text{use the theorem 6}}{\text{.8}}{\text{.3 }}\left( {{\text{ see page 476}}} \right){\text{ }} \cr
& \frac{d}{{dx}}\left[ {\tanh u} \right] = {\operatorname{sech} ^2}u\frac{{du}}{{dx}} \cr
& y' = {x^3}\left( {2\tanh \sqrt x } \right)\left( {{{\operatorname{sech} }^2}\sqrt x } \right)\left( {\sqrt x } \right)' + {\tanh ^2}\left( {\sqrt x } \right)\left( {{x^3}} \right)' \cr
& y' = {x^3}\left( {2\tanh \sqrt x } \right)\left( {{{\operatorname{sech} }^2}\sqrt x } \right)\left( {\frac{1}{{2\sqrt x }}} \right) + {\tanh ^2}\left( {\sqrt x } \right)\left( {3{x^2}} \right) \cr
& {\text{simplifying}} \cr
& y' = {x^{5/2}}\tanh \sqrt x {\operatorname{sech} ^2}\sqrt x + 3{x^2}{\tanh ^2}\sqrt x \cr} $$
