91Ó°ÊÓ

We have given expression:$f(x,y,z)=\frac{x{y}^2}{x+y-z}$

Consider the function $g:f^2→f$.Then the domain of the function is

$Domain\left(g\right)=\left\{\left(x,y,z\right)∈f^2:g(x,y,z)isdefined\right\}$

Since the rational function $f(x,y,z)=\frac{x{y}^2}{x+y-z}$is defined for all role="math" localid="1653315545139" $\left(x,y,z\right)∈f^2$such that

$$\begin{gathered} x+y-z≠0 \\ x+y≠z \end{gathered}$$

The domain of the function is$Domain\left(f\right)=\left\{\left(x,y,z\right)∈f^3:x+y≠z\right\}$

Since $x{y}^2$and $x+y-z$ being a polynomial function of two variable is continuous for every point on $f^2$.

The rational function is continuous where all those points where role="math" localid="1653316004175" $f(x,y,z)=\frac{x{y}^2}{x+y-z}$defined.

The rational function is discontinuous only at the points where role="math" localid="1653315941308" $x+y-z=0$that is $x+y=z$.

Hence, the given function$f(x,y,z)=\frac{x{y}^2}{x+y-z}$ is continuous on the set$\left\{\left(x,y,z\right)∈f^3:x+y≠z\right\}$.