Answer
No, the function $ f\left( x \right)=\frac{{{x}^{2}}+8x}{{{x}^{2}}-8x}$ is not continuous at $0$.
Work Step by Step
Consider the function $ f\left( x \right)=\frac{{{x}^{2}}+8x}{{{x}^{2}}-8x}$,
First check whether the function is defined at the point $ a $ or not.
Find the value of $ f\left( x \right)$ at $ a=0$,
$ f\left( 0 \right)=\frac{{{\left( 0 \right)}^{2}}+8\left( 0 \right)}{{{\left( 0 \right)}^{2}}-8\left( 0 \right)}$
The function is not defined at the point $0$.
Thus, the function does not satisfy the first property of being continuous.
Hence, the function $ f\left( x \right)=\frac{{{x}^{2}}+8x}{{{x}^{2}}-8x}$ is not continuous at $0$.
