Answer
$\bf True$.
Work Step by Step
$\bf True$.
$f(x)=y=\frac{lnx}{x}$
$y'=\frac{1-lnx}{x^{2}}$
Now, $x^{2}y'+xy=x^{2}\frac{1-lnx}{x^{2}}+x\frac{lnx}{x}$
$=1-lnx+lnc$
$=1$
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