This one is slightly tricky, but you can apply Rouché directly.
Let $g(z) = 3z^2 + 1$. Note that $|g(z)| \geq 2$ for $|z| = 1$ with equality only for $z = \pm i$ (because $g$ maps the unit circle onto the circle with radius $3$ centered at $1$).
On the other hand for all $|z| = 1$ we have the estimate $h(z) = |f(z) - g(z)| = |z(z^3 + 1)| \leq 2$ and for $z = \pm i$ we have $h(\pm i) = \sqrt{2} < 2 \leq |g(\pm i)|$. Therefore $|f(z) - g(z)| < |g(z)|$ for all $|z| = 1$ and thus Rouché can be applied.
Choose $g(z) = z^5 + 1$. The zeros of $g$ are $e^{k\pi i/5},\; k \in \{1,3,5,7,9\}$, two of which lie in the right half plane.
Then a slightly generalised version of Rouché's theorem tells you that $h$ also has two zeros in the right half plane.
The standard formulation of Rouché's theorem demands that $\lvert h-g\rvert < \lvert h\rvert$ (or $\lvert h-g\rvert < \lvert g\rvert$) on the boundary $\partial V$ of the region, but looking at the proof, whose main part is the observation that the number of zeros of $f_\lambda = g + \lambda(h-g)$ in $V$,
$$N_\lambda = \frac{1}{2\pi i}\int_{\partial V} \frac{g'(z) + \lambda\bigl(h'(z) - g'(z)\bigr)}{g(z) + \lambda\bigl(h(z) - g(z)\bigr)}\, dz$$
is independent of $\lambda \in [0,\,1]$, since no $f_\lambda$ has a zero on $\partial V$.
The role of the condition $\lvert h-g\rvert < \lvert g\rvert (\text{ or } \lvert h\rvert)$ on $\partial V$ is solely to guarantee that $g + \lambda(h-g)$ also has no zeros on $\partial V$. If we can determine that in a different way, the conclusion still holds.
For a large enough semicircle in the right half plane, we have $\lvert 5z^3 + 2z^2 + 4z\rvert < \lvert z^5 + 1\rvert$ since the RHS is a polynomial of higher degree than the LHS.
On the imaginary axis, although $\lvert g-h\rvert > \lvert g\rvert$ on some parts, expanding shows
$$g(it) + \lambda\bigl(h(it) - g(it)\bigr) = 1 + it^5 + \lambda(-5it^3 - 2t^2 + 4it) = (1-2\lambda t^2) + it(t^4 - 5\lambda t^2 + 4\lambda),$$
and the real part vanishes only for $\lambda = \dfrac{1}{2t^2}$ (with $t^2 \geqslant \frac12$, since $\lambda \leqslant 1$), when the imaginary part is
$$t\left(t^4 - \frac{5}{2} + \frac{2}{t^2}\right) = \frac1t\left(t^6 - \frac{5}{2}t^2 + 2\right),$$
and the polynomial $x^3 - \frac52x + 2$ has no positive real roots, so we conclude $g(z) + \lambda(h(z)-g(z)) \neq 0$ on the imaginary axis.
Best Answer
Concerning the first part:
You want to prove that there is no real root to the polynomial $p(z)=z^6+9z^4+z^3+2z+4$.
Now, the idea is to group terms to prove that $p(z)>0$ for all real $z$.
First of all, you can get away that $z^3$ term using $z^6+z^3+\frac{1}{4} \ge 0$ which is equivalent to $(z^3+\frac{1}{2})^2 \ge 0$ which holds for all real $z$.
Now, it suffices to prove $9z^4+2z+4 \ge 0$.
For $z>0$ this is clearly true. Assume $z<0$. Then we want to prove that $9x^4+4 \ge 2x$ holds for all $x>0$. But by the AM-GM inequality you have $9x^4+3=4 \cdot \frac{9x^4+1+1+1}{4} \ge 4 \cdot \sqrt[4]{9x^4 \cdot 1 \cdot 1 \cdot 1}=4x \cdot \sqrt{3}>4x>2x$. So $9x^4+4 >2x+1>2x$ holds for all positive $x$.
Hence $p(z)$ has no real roots i.e. there must be one root in each the second and the third quadrant.