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 $x \ge 1$.

For the second portion, set 
$$
x(x^5+3x^3-3)=0
$$
Which yields $x=0$ or $x^5+3x^3-3$. If $x=0$, then the strict inequality does not hold and clearly if $x<0$, the product is positive.
Furthermore, checking the derivative of  $x^5+3x^3-3$, we see that the function is strictly increasing, hence the zero of the odd function is unique. Then, that the zero occurs in between $0$ and $1$ follows form the intermediate value theorem. Again, you can eliminate options larger than 1 since the function is increasing.
