As a follow-up, we are given:
$$f(x)=x^6\ln(x)$$
The first thing I would observe is that the domain of this function is $0<x$.
Using the product rule, we find:
$$f'(x)=\left(6x^5\right)\left(\ln(x)\right)+\left(x^6\right)\left(\frac{1}{x}\right)=x^5\left(6\ln(x)+1\right)$$
Using the product rule again, we find:
$$f''(x)=\left(5x^4\right)\left(6\ln(x)+1\right)+\left(x^5\right)\left(\frac{6}{x}\right)=x^4\left(30\ln(x)+11\right)$$
As $0<x$, we need only look at the critical value that comes from:
$$30\ln(x)+11=0\implies x=e^{\Large{-\frac{11}{30}}}$$
Using test values on either side of this critical value, such as $e^{-1}$ or $e$, we see that we have:
$$30\ln(x)+11<0$$ on $\left(0,e^{\Large{-\frac{11}{30}}}\right)$ and thus $f$ is concave down on this interval.
$$30\ln(x)+11>0$$ on $\left(e^{\Large{-\frac{11}{30}}},\infty\right)$ and thus $f$ is concave up on this interval.
Thus, we have a point of inflection at:
$$\left(e^{\Large{-\frac{11}{30}}},f\left(e^{\Large{-\frac{11}{30}}}\right)\right)=\left(e^{\Large{-\frac{11}{30}}},-\frac{11}{30}e^{\Large{-\frac{11}{5}}}\right)$$
Here is a plot of the given function with the point of inflection shown:
View attachment 7538