Answer
$y=x-1$
Work Step by Step
$y=x^{2} \ln x\\
y^{\prime}=\left(x^{2}\right)\left(\frac{1}{x}\right)+(\ln x)(2 x)\\
y^{\prime}=x+2 x \ln x\\
y^{\prime}=x(1+2 \ln x)$
Gradient at point $(1,0)$
$y^{\prime}=1(1+2 \ln 1)\\
y^{\prime}=1$
Tangent line at point $(1,0)$
$y-0=1(x-1)\\
y=x-1$
