The equation for the ellipse states $x^2+y^2-2x=0$ (...) from which I have $\rho(\theta)= \displaystyle\frac{\cos\theta}{1+\cos^2\theta}$ with $0\leq\theta\leq\pi$.
The equation of the ellipse is
\begin{equation*}
2x^{2}+y^{2}-2x=0
\end{equation*}
from which I've obtained
\begin{equation*}
\rho (\theta)=\frac{2\cos \theta }{\cos ^{2}\theta +1},\qquad \text{with }-\pi/2 \leq
\theta <\pi/2,
\end{equation*}
because the tangent to the ellipse at $(x,y)=(0,0)$ is the vertical line $x=0$. With these corrections the volume integral becomes
\begin{eqnarray*}
V &=&\int_{-\pi /2}^{\pi /2}\left( \int_{0}^{\frac{2\cos \theta }{\cos
^{2}\theta +1}}2\rho \sqrt{1-\rho ^{2}}d\rho \right) \,d\theta \\
&=&-\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left. \left( 1-\rho ^{2}\right)
^{3/2}\right\vert _{0}^{\frac{2\cos \theta }{\cos ^{2}\theta +1}}\,d\theta
\\
&=&-\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( 1-\left( \frac{2\cos \theta }{
\cos ^{2}\theta +1}\right) ^{2}\right) ^{3/2}-1\,\ d\theta
\end{eqnarray*}
Simplifying, we have that
\begin{eqnarray*}
V &=&\frac{2}{3}\pi -\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( \left( \frac{
\sin ^{2}\theta }{\cos ^{2}\theta +1}\right) ^{2}\right) ^{3/2}\,d\theta \\
&=&\frac{2}{3}\pi -\frac{2}{3}\int_{-\pi /2}^{\pi /2}\left( \frac{\sin
^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta \\
&=&\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin ^{2}\theta
}{\cos ^{2}\theta +1}\right) ^{3}\,d\theta ,
\end{eqnarray*}
because the integrand is an even function. The remaining integral can be
converted into a rational function of $t=\tan \frac{\theta }{2}$ by the
Weirstrass substitution $t=\tan \frac{
\theta }{2}$, which I then evaluated in SWP (Scientific Work
Place):
\begin{eqnarray*}
V &=&\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin
^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta ,\qquad t=\tan \frac{
\theta }{2} \\
&=&\frac{2}{3}\pi -\frac{64}{3}\int_{0}^{1}\frac{t^{6}}{\left(
1+t^{4}\right) ^{3}\left( 1+t^{2}\right) }\,dt \\
&=&\cdots \\
&=&\frac{4}{3}\pi -\frac{7}{12}\sqrt{2}\pi =\frac{16-7\sqrt{2}}{12}\pi .
\end{eqnarray*}
ADDED. The integrand in $t$ can be expanded into partial fractions as follows
\begin{equation*}
\frac{t^{6}}{\left( 1+t^{4}\right) ^{3}\left( 1+t^{2}\right) }=\frac{1}{8}
\frac{t^{2}-1}{1+t^{4}}+\frac{1}{4}\frac{3+t^{2}}{\left( 1+t^{4}\right) ^{2}}
-\frac{1}{2}\frac{1+t^{2}}{\left( 1+t^{4}\right) ^{3}}-\frac{1}{8\left(
1+t^{2}\right) }.
\end{equation*}
ADDED 2. WolframAlpha computation of $V =\frac{2}{3}\pi -\frac{4}{3}\int_{0}^{\pi /2}\left( \frac{\sin^{2}\theta }{\cos ^{2}\theta +1}\right) ^{3}\,d\theta$ confirms the result above.
I think your method is correct (of converting first to cylindrical, and then to spherical), but you did make one mistake. Here I will convert directly to spherical from Cartesian using the transformation:
$$
\begin{align*}
x &= \rho \sin \phi \cos \theta \\
y &= \rho \sin \phi \sin \theta \\
z &= \rho \cos \phi
\end {align*}
$$
So the equation $y = (x^2+y^2)^2 + z^4$ becomes:
$$ \rho \sin \phi \sin \theta = \rho^4 \left( \sin^4 \phi + \cos^4 \phi \right) $$
Which you can solve for $\rho$:
$$ \rho = \sqrt[3]{\frac{\sin \phi \sin \theta}{\sin^4\phi + \cos^4\phi}} $$
Note it is $\sin\phi\sin\theta$ in the numerator, and not $\sin^2\theta$.
Now to compute the volume, you should be able to integrate the volume differential $(\rho^2 \sin\phi)$ over the region where $\rho$ is between 0 and the expression above (and the appropriate bounds for $\theta$ and $\phi$).
Best Answer
We have
$$\iiint\limits_{\text{Solid}}{\text dx\text dy\text dz}=\int_{-1}^1\, dz\int_0^{\sqrt{1-z^2}} \, dx\int_{z^2}^{2-2x^2-z^2}\, dy=\frac{\pi}2$$
and by
$x=\rho \sin \theta$
$z=\rho \cos \theta$
$y=y$
$$\int_0^\pi\text d\theta\displaystyle\int_0^1\rho\text d\rho\displaystyle\int_{\rho^2\cos^2\theta}^{2-\rho^2-\rho^2\sin^2\theta}\text dy=\frac{\pi}2$$
The result is the same but in your set up $x$ and $z$ have been exchanged.