equation of tangent at x = 3:
y − y₁ = m(x − x₁)

equation of tangent at x = 3:
y − 3 = 1(x − 3)
y = x − 3 + 3
y = x

To find another point with the same slope, consider the symmetry of the curve.

m = y '(3)

$\text{since }y= 3\sqrt[3]{x-2}=3(x-2)^{\frac{1}{3}}$

$\text{so }y'= (x-2)^{-\frac{2}{3}}= \frac{1}{(x-2)^{\frac{2}{3}}}$

and y '(3) = 1 = m

x₁ = 3

$y_{1} = 3\sqrt[3]{(3)-2}=3$