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 ?

Notice, we have $$f(x)=\frac{1}{x^3}-x^3$$
$$\frac{d}{dx}(f(x))=\frac{d}{dx}\left(\frac{1}{x^3}-x^3\right)$$
$$f'(x)=\frac{-3}{x^4}-3x^2$$
$$f'(x)=-3\left(\frac{1}{x^4}+x^2\right)$$
$$f'(x)=-3\left(\left(\frac{1}{x^2}\right)^2+x^2\right)\tag 1$$
We know that $$\left(\frac{1}{x^2}\right)^2> 0\ \ \ \ \forall \ x>0 \tag 2$$ & $$\left(x\right)^2> 0\ \ \ \ \forall \ x>0 \tag 3$$ From (2) & (3) we get $$\left(\frac{1}{x^2}\right)^2+x^2> 0$$$$\iff -3\left(\left(\frac{1}{x^2}\right)^2+x^2\right)<0\ \ \ \ \forall \  \ x>0\tag 4$$ Hence, from (1) & (4), we find that $$f'(x)<0\ \ \ \forall \ \ \ x>0$$  
Hence, the function $f(x)  $ is decreasing for all $x>0$
