Step 1: Condition to test:
The point must lie on the curve, and the normal must have slope -1, which means the tangent slope $\frac{x^2}{6y}$ must equal 1.
Step 2: Test option (B):
At $\left(4, \frac83\right)$: $9y^2 = 9\cdot\frac{64}{9} = 64 = 4^3$, so it lies on the curve. Slope of tangent $= \frac{x^2}{6y} = \frac{16}{16} = 1$, so the normal has slope -1 and equal intercepts.
Options (A), (C), (D) have $x = -4$, which gives $x^3 = -64 < 0$, but $9y^2$ is never negative. So only (B) works.
Final Answer:
$\left(4, \frac83\right)$.
\[ \boxed{\left(4,\frac{8}{3}\right)} \]