Prove function is < 0 
For all $x$ if $x^6 + 3x^4 - 3x < 0$ then $0 < x < 1$. Prove this.
(1) Find the negation
(2) Prove

(1) The negation is simply, $\exists x$, $x^6 + 3x^4 - 3x < 0 \wedge (x \le 0 \vee x \ge 1)$
(2) The proof is the difficult part here.
We prove the contrapositive. It is easy to prove it for the condition that $x \le 0$ but it is harder for $x \ge 1$.

If $x <-1$, then $f (x)>f (-x)$ so it will suffice to check the cases for $x>1$. Notice that $x^6>x,$ since $(x\cdot x\dots x>x\cdot 1\dots 1=x $. Can you finish with the same argument for $3x^4$ and $3x $? What can you conclude for values of $f(x) $ outside of the unit interval?
