If k is even number then for what value of x and y, x2k−(xy)k+y2k=0

Suppose y is zero, then the only solution to ${\displaystyle x^{2k}=0\text{is}x=0}$. Thus, ${\displaystyle x=y=0}$ is one of the possible solutions.
Now assume y is not zero. Then dividing everything by ${\displaystyle y^{2k}}$ you get
${\displaystyle {\left(\frac{x}{y}\right)}^{2k}-{\left(\frac{x}{y}\right)}^k+1=0}$
or, after substituting ${\displaystyle z={\left(\frac{x}{y}\right)}^k}$,
${\displaystyle z^2-z+1=0}$
This quadratic equation has no real solutions.
This means ${\displaystyle x=y=0}$ is the only possible solution, and even that for ${\displaystyle k>0}$ because when ${\displaystyle k=0}$ the expression ${\displaystyle 0^0}$ is not well-defined.