Answer is: option3

$\frac{3e}{10}$

Solution:

$y=x^2\ln x$

We need to find $dy$ when $x=e$ and $dx=0.1$.

Differentiating both sides with respect to $x$:

$\frac{dy}{dx}=\frac{d}{dx}(x^2\ln x)$.

Using the product rule, where $u=x^2$ and $v=\ln x$:

$\frac{d}{dx}(x^2\ln x)=x^2\cdot \frac{d}{dx}(\ln x)+\ln x\cdot \frac{d}{dx}(x^2)$.

$=x^2\cdot \frac{1}{x}+\ln x\cdot (2x)$.

$=x+2x\ln x$.

Thus,

$dy=(x+2x\ln x)dx$.

Substituting $x=e$:

$dy=(e+2e\ln e)dx$.

Since $\ln e=1$, this simplifies to:

$dy=(e+2e\cdot 1)dx$.

$dy=(e+2e)dx$.

$dy=3e\cdot dx$.

Substituting $dx=0.1$:

$dy=3e\times 0.1$.

$dy=\frac{3e}{10}$.

The answer is $\frac{3e}{10}$.