The function $f(x)=\log_e(x^3+\sqrt{(x^6+1)})$ is of the following types:

odd

$f(x)=\log_e(x^3+\sqrt{x^6+1})⇒f(-x)=\log(-x^3+\sqrt{x^6+1})$

so $f(-x)=\log((-x^3+\sqrt{x^6+1})×\frac{(\sqrt{x^6+1}+x^3)}{(\sqrt{x^6+1}+x^3)})$

$f(-x)=\log(\frac{1}{\sqrt{x^6+1}+x^3)})=-\log(x^3+\sqrt{x^6+1})$

$⇒f(-x)=-f(x)$ → odd function.

Now, $f(x)=\log(x^3+\sqrt{(x^6+1)})$

$f'(x)=\frac{1}{x^3+\sqrt{x^6+1}}\left(3x^2+\frac{6x^2}{2\sqrt{x^6+1}}\right)=\frac{3x^2}{\sqrt{x^6+1}}>0$ ⇒ f(x) is increasing.