Answer
$\dfrac{\ln (2)}{3}$
Work Step by Step
Integrate the integral first with respect to $x$, then $z$, and then $y$
$ \int_{0}^1 \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} \dfrac{z}{y+1} dx dz dy= \int_{0}^1 \int_{0}^{1} [\dfrac{xz}{y+1}]_{0}^{\sqrt{1-z^2}} dz dy=\int_{0}^1 \dfrac{1}{y+1} dy \int_{0}^{1} z\sqrt{1-z^2} dz$
Suppose $1-z^2 =a\implies z dz=-\dfrac{da}{2}$
Now, we have
$\int_{0}^1 \dfrac{1}{y+1} dy \int_{0}^{1} z\sqrt{1-z^2} dz=[\ln |y+1|]_{0}^1 \int_{0}^{1} \sqrt a [-\dfrac{1}{2}] da$
Hence, $\int_{0}^1 \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} \dfrac{z}{y+1} dx dz dy=(\dfrac{-1}{2}) (\ln 2)\int_0^1 \sqrt {a} da=\dfrac{\ln 2}{3}$
