I'm going to take a wild guess and assume you mean
[tex]f(x,y,z)=z(x^2+y^2+z^2)^{-3/2}=\dfrac z{(x^2+y^2+z^2)^{3/2}}[/tex]
(as opposed to subtracting 3/2, say)
Convert to spherical coordinates, taking
[tex]\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]
Now when [tex]z=\dfrac72[/tex], below the "cap" of the ball we have a right triangle with hypotenuse coinciding with the ball's radius and a leg of length [tex]\dfrac72[/tex], with the angle between them [tex]\varphi[/tex] satisfying
[tex]\cos\varphi=\dfrac{\frac72}7=\dfrac12\implies\varphi=\cos^{-1}\dfrac12=\dfrac\pi3[/tex]
Furthermore, when [tex]z=\dfrac72[/tex], we have
[tex]\dfrac72=\rho\cos\varphi\implies\rho=\dfrac72\sec\varphi[/tex]
Call the cap [tex]C[/tex]. Then the triple integral over [tex]C[/tex] is
[tex]\displaystyle\iiint_Cf(x,y,z)\,\mathrm dV=\int_{\varphi=0}^{\varphi=\pi/3}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=7/2\,\sec\varphi}^{\rho=7}\frac{\rho\cos\varphi}{(\rho^2)^{3/2}}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]
[tex]=\displaystyle2\pi\int_{\varphi=0}^{\varphi=\pi/3}\int_{\rho=7/2\,\sec\varphi}^{\rho=7}\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi[/tex]
[tex]=\displaystyle\pi\int_{\varphi=0}^{\varphi=\pi/3}\sin2\varphi\left(7-\dfrac72\sec\varphi\right)\,\mathrm d\varphi[/tex]
[tex]=\displaystyle\frac{7\pi}2\int_{\varphi=0}^{\varphi=\pi/3}(2\sin2\varphi-\sin2\varphi\sec\varphi)\,\mathrm d\varphi=\frac{7\pi}4[/tex]
