I need to find the surface area of 2($x^2$ + $y^2$) = $y$ cut by $4z^2$ = $1-2y$.
I did the following:From first equation, I got $x$ = $+$ $-$ $\frac{y-y^2}{2}$
I then did partial derivative of $x$ with respect to $y$ and $z$.
I took $x$ positive because I assume that I will get the same things if i take $x$ negative.
After setting up my integral, and calculating, I get solution $\frac{1}{3}$.
Is this good?
Edit:$$\int_0^\frac{1}{2} \int_0^{\frac{1}{2}-2z^2} \sqrt\frac{12y^2-6y+1}{2(y-2y^2)} dydz$$
I realized now that maybe it is easier if I expressed $z$ from the 2nd equation.
Continuing with your approach, we let $x(y,z) = \sqrt{\dfrac y2-y^2}$ so that
$$\sqrt{1+\left(\frac{\partial x}{\partial y}\right)^2+\left(\frac{\partial x}{\partial z}\right)^2} = \frac1{2\sqrt{2y-4y^2}}$$
Project the part of the cylindrical cutout (denoted $S$) with $x>0$ onto the $y,z$ plane, which gives the set
$$S'=\left\{(0,y,z) \mid z\in\left[-\frac12,\frac12\right] \land y\in\left[0,\frac12-2z^2\right]\right\}$$
By symmetry, the integral over $S$ is twice the integral over $S'$, and twice again due to symmetry in the $z$-integral, hence
$$\iint_S dA = \sqrt2 \int_0^{\tfrac12} \int_0^{\tfrac12-2z^2} \frac{dy\,dz}{\sqrt{y-2y^2}}$$
Alternatively, another way we can parameterize $S$ is in cylindrical coordinates with $(x,y,z)=\left(\frac14\cos\theta,\frac14\sin\theta+1,z\right)=:\vec s(\theta,z)$, so that
$$S = \left\{(\theta,z) \mid \theta\in[0,2\pi] \land z\in\left[-\frac{\sqrt{1-\sin\theta}}{2\sqrt2},\frac{\sqrt{1-\sin\theta}}{2\sqrt2}\right]\right\}$$
Take the normal to $S$ to be $\vec n = \pm\dfrac{\partial\vec s}{\partial\theta} \times \dfrac{\partial\vec s}{\partial z}$. Putting everything together and taking advantage of symmetry again, (you can confirm that) we get the same result with
$$\iint_S dA = \frac14 \int_0^{2\pi} \int_{-\tfrac{\sqrt{1-\sin\theta}}{2\sqrt2}}^{\tfrac{\sqrt{1-\sin\theta}}{2\sqrt2}} dz \, d\theta = \int_0^\pi \int_0^{\tfrac{\sqrt{1-\cos\theta}}{2\sqrt2}} dz \, d\theta$$