Answer
$f(x,y,z)$ is continuous for $y \geq x^2$ and $z \gt 0$
Work Step by Step
We are given that $f(x,y,z)=\sqrt {y-x^2} \ln z$
The function $f(x,y,z)=\sqrt {y-x^2} \ln z$ represents a squared root function which cannot be less than $0$. Also, $\ln$ can only be taken of a positive number.
Thus,
$ {y-x^2} \geq 0$
or, $y \geq x^2 $
and $z \gt 0$
This means that $ y \geq x^2 ; z \gt 0$
Hence, $f(x,y,z)$ is continuous for $y \geq x^2$ and $z \gt 0$
