Answer
$$\frac{{{d^3}y}}{{d{x^3}}} = \frac{2}{x}$$
Work Step by Step
$$\eqalign{
& {\text{let }}y = {x^2}\ln x \cr
& {\text{find }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^2}\ln x} \right] \cr
& {\text{using the product rule }} \cr
& \frac{{dy}}{{dx}} = {x^2}\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& \frac{{dy}}{{dx}} = {x^2}\left( {\frac{1}{x}} \right) + \ln x\left( {2x} \right) \cr
& \frac{{dy}}{{dx}} = x + 2x\ln x \cr
& \cr
& {\text{find }}\frac{{{d^2}y}}{{d{x^2}}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {\frac{{dy}}{{dx}}} \right] \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {x + 2x\ln x} \right] \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left[ x \right] + \frac{d}{{dx}}\left[ {2x\ln x} \right] \cr
& {\text{product rule}} \cr
& \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left[ x \right] + 2x\frac{d}{{dx}}\left[ {\ln x} \right] + 2\ln x\frac{d}{{dx}}\left[ x \right] \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 1 + 2x\left( {\frac{1}{x}} \right) + 2\ln x\left( 1 \right) \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 1 + 2 + 2\ln x \cr
& \frac{{{d^2}y}}{{d{x^2}}} = 3 + 2\ln x \cr
& \cr
& {\text{find }}\frac{{{d^3}y}}{{d{x^3}}} \cr
& \frac{{{d^3}y}}{{d{x^3}}} = \frac{d}{{dx}}\left[ {\frac{{{d^2}y}}{{d{x^2}}}} \right] \cr
& \frac{{{d^3}y}}{{d{x^3}}} = \frac{d}{{dx}}\left[ {3 + 2\ln x} \right] \cr
& \frac{{{d^3}y}}{{d{x^3}}} = \frac{d}{{dx}}\left[ 3 \right] + \frac{d}{{dx}}\left[ {2\ln x} \right] \cr
& \frac{{{d^3}y}}{{d{x^3}}} = 1 + 2\left( {\frac{1}{x}} \right) \cr
& \frac{{{d^3}y}}{{d{x^3}}} = \frac{2}{x} \cr} $$
