Find the equation of the tangent plane to the surface given by $$x^{\frac{2}{3}}+y^{\frac{2}{3}}+z^{\frac{2}{3}}=2^{\frac{2}{3}}$$ at the given point $(x_0,y_0,z_0)$
Then find the intersection of the tangent curve with the $x,y,z$ axis,show that the sum of the squares of these distances from the origin is a constant.
considering the given point, the equation transforms to $$\frac{2}{3}x_0^{\frac{-1}{3}}\left(x-x_{0}\right)+\frac{2}{3}y_0^{\frac{-1}{3}}\left(y-y_{0}\right)+\frac{2}{3}z_0^{\frac{-1}{3}}\left(z-z_{0}\right)=0$$
As I know the intersection of the tangent surface with the $x,y,z$-axis is the set of all points such that (resp):
$$x_0^{\frac{-1}{3}}\left(x-x_{0}\right)-y_0^{\frac{-1}{3}}y_{0}-z_0^{\frac{-1}{3}}z_{0}=0$$ $$x_0^{\frac{-1}{3}}x_{0}-y_0^{\frac{-1}{3}}(y-y_{0})-z_0^{\frac{-1}{3}}z_{0}=0$$ $$x_0^{\frac{-1}{3}}x_0-y_0^{\frac{-1}{3}}y_{0}-z_0^{\frac{-1}{3}}(z-z_{0})=0$$
I don't know what else should I do.
Take the first one (intersection with x-axis) for example,
$x_0^{-1/3}\left(x-x_{0}\right)-y_0^{-1/3}y_{0}-z_0^{-1/3}z_{0}=0$
$x_0^{-1/3} x - (x_0^{2/3} + y_0^{2/3} + z_0^{2/3}) = 0$
As $(x_0, y_0, z_0)$ is on the plane, $x_0^{2/3} + y_0^{2/3} + z_0^{2/3} = 2^{2/3}$
So $x = 2^{2/3} x_0^{1/3}$ and if point $A$ is intersection with x-axis, its coordinates is $(2^{2/3} x_0^{1/3}, 0, 0)$ and the distance from the origin is $d_A = 2^{2/3} x_0^{1/3}$.
Similarly, $d_B = 2^{2/3} y_0^{1/3}, d_C = 2^{2/3} z_0^{1/3}$
Squaring and adding their distances from the origin, we get
$d^2 = d_A^2 + d_B^2 + d_C^2 = 2^{4/3} (x_0^{2/3} + y_0^{2/3} + z_0^{2/3}) = 4$ which is a constant.