Answer
See explanation.
Work Step by Step
Let $w=\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)^{k}=\left[\sum_{i=1}^{n} x_{i}^{2}\right]^{k}, n>2 .$ To ilustrate the
We determine the partial derivative with respect to $x_{1}$
\[
\begin{aligned}
\frac{d w}{\partial x_{1}} &=k\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)^{k-1} \frac{\partial}{\partial x_{1}}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) \\
&=2 k x_{1}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)^{k-1}
\end{aligned}
\]
Analogously for a general case, we have:
\[
\begin{aligned}
\frac{d w}{\partial x_{i}} &=k\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)^{k-1} \frac{\partial}{\partial x_{i}}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) \\
&=2 k x_{i}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)^{k-1} \quad i=1,2,3, \cdots n
\end{aligned}
\]
We repeat this proccedure using second derivtives
\[
\begin{aligned}
&\left(\frac{\partial w}{\partial x_{1}}\right)\frac{\partial}{\partial x_{1}}=\frac{\partial^{2} w}{\partial x_{1}^{2}} \\
&=2 k\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)^{k-1}+4 k(k-1) x_{1}^{2}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)^{k-2} \\
&=2 k\left(x_{1}^{2}+\cdots+x_{n}^{2}\right)^{k-2}\left[x_{1}^{2}+\cdots+x_{n}^{2}+2(k-1) x_{1}^{2}\right]
\end{aligned}
\]
Generalizing this
\[
\begin{aligned}
\frac{\partial^{2} w}{\partial x_{i}^{2}} &=\frac{\partial}{\partial x_{i}}\left(\frac{\partial w}{\partial x_{1}}\right) \\
&=2 k\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)^{k-1}+4 k(k-1) x_{i}^{2}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)^{k-2} \\
&=2 k\left(x_{1}^{2}+\cdots+x_{n}^{2}\right)^{k-2}\left[x_{1}^{2} \cdots+2 k x_{i}^{2}+\cdots+x_{n}^{2}\right]
\end{aligned}
\]
In the sum of the n terms
\[
0=\sum_{i=1}^{n} \frac{\partial^{2} w}{\partial x_{i}^{2}}
\]
\[
\begin{aligned}
\Rightarrow & 2 x\left(x_{1}^{2}+\cdots+x_{n}^{2}\right)^{n-2}+\left(x_{1}^{2}+\cdots+x_{n}^{2}\right)(n-1+2 k)=0 \\
& \Rightarrow 2 k(2 k+n-1)\left(x^{2}+\cdots+x_{n}^{2}\right)^{k-1}=0 \\
& \Rightarrow 0=-1+2 k+n \quad \text { or } \quad 0=2 k
\end{aligned}
\]
