Show that $f$ is a decreasing function It's given that $f(x)=\frac{1}{x^3}-x^3$ for $x>0$ show that $f$ is a decreasing function. 
My attempt $f'(x)=-3x^{-4}-3x^2$
$x^6=-1$
How to continue by my attempt ?

you need to show $f'$ is always negative.
suppose $f'$ is positive. then $-3x^{-4}-3x^2=k>0$
$-3-3x^6=kx^4$
$3x^6+kx^4+3=0$
since $x^2$ is always greater than or equal to zero, so it $3(x^2)^3+k(x^2)^2$, so if you add $3$ the result is always positive. thus no such $k$ exists
