Lagrange to find the maximum. I do not know how to get the following:
the answer is 1/n or 1/n^n But I can't get that result.
We have to use Lagrange theorem $\triangledown f = \lambda \triangledown g$ and I got the gradiants:

$\triangledown f = 2x{_{k}}\prod x{_{j}}^{2}$

and

$\triangledown g = 2x{_{k}}$

but than I can't get the correct result. Please help me.
With

$f(x_1, x_2, \ldots, x_n) = \prod_1^n x_i^2 \tag{1}$

and

$g(x_1, x_2, \ldots, x_n) = \sum_1^n x_i^2 = 1, \tag{2}$

we have, as pointed out by our OP SoHCahToha,

$(\nabla f)_j = 2x_j \prod_{1, i \ne j}^n x_i^2, \tag{3}$

and

$(\nabla g)_j = 2x_j.  \tag{4}$

With Lagrange multiplier $\lambda$, we conclude that

$(\nabla f)_j = \lambda (\nabla g)_j \tag{5}$

for all $j$, which by (3)-(4) becomes

$2x_j \prod_{1, i \ne j}^n x_i^2 = \lambda (2x_j). \tag{6}$

We note that

$f(x_1, x_2, \ldots, x_n) \ge 0 \tag{7}$

for all vectors $(x_1, x_2, \ldots, x_n)$, and 

$f(x_1, x_2, \ldots, x_n) = 0 \tag{8}$

only when $x_i = 0$ for at least one index $i$.  Therefore, the maximum of $f$ occurs at a point where $x_i \ne 0$ for all $i$, so division of (6) by $2x_j$ is permissible, leading to

$\prod_{1, i \ne j}^n x_i^2 = \lambda \tag{9}$

for every $j$.  Furthermore, from (6) we see that $\lambda \ne 0$ since no $x_j = 0$ at the maximum, so we can form the quotients

$\dfrac{\prod_{1, i \ne j}^n x_i^2}{\prod_{1, i \ne k}^n x_i^2} = 1 \tag{10}$

for every pair of indices $j, k$.  (10) is trivial when $j = k$, but for $j \ne k$ becomes

$\dfrac{x_k^2}{x_j^2} = 1 \tag{11}$

or

$x_k^2 = x_j^2; \tag{12}$

using (2) and (12), we see that

$n x_j^2 = 1 \tag{13}$

or

$x_j^2 = \dfrac{1}{n}; \tag{14}$

thus, for every $j$,

$x_j = \pm \dfrac{1}{\sqrt n}. \tag{15}$

It then follows that

$f(x_1, x_2, \ldots, x_n) = \prod_1^n \dfrac{1}{n} = \dfrac{1}{n^n} = n^{-n} \tag{16}$ 

is the maximum possible value of $f(x_1, x_2, \ldots, x_n)$ subject to (2).

Technically, we need to prove the points whose coordinates are given by (15) are in fact true maxima and not some other form of extrema such as saddle points.  But since $f(x_1, x_2, \ldots, x_n)$ is a smooth function on the compact sphere given by (2), we know it takes its maximal value, where the derivatives of $f$ vanish.  But we have seen that at every point where $\nabla f = 0$, $f$ takes the value $n^{-n}$; thus this must in fact be the greatest possible value of $f$ subject to (2).
This is very simple, and equivalent to:

$$
\mathcal{P}: \max f=y^{n}\\
 ny=1, y\ge 0\\
y=x_i^2 \ \forall i
$$

The solution is then 
$$
x_i=\sqrt{1/n} \ \forall i
$$