Answer
$\frac{28}{3}$
Work Step by Step
Step 1. Given the equation $y^2=x^3$, differentiate both sides with respect to $x$: $2yy'=3x^2$, which gives $y'=\frac{3x^2}{2y}$,
Step 2. The slope of the tangent line at a point $(x_0,y_0)$ is $m=y'=\frac{3x_0^2}{2y_0}$
Step 3. The slope of the line $y=-x/3+b$ is $n=-1/3$; letting $mn=-1$, we have $\frac{3x_0^2}{2y_0}=3$ or $y_0=x_0^2/2$
Step 4. Since $(x_0,y_0)$ is on the curve, we have $(x_0^2/2)^2=x_0^3$, which gives $x_0=0, 4$ and $y_0=0, 8$
Step 5. Discard point $(0,0)$. For point $(4,8)$ on the line, we have $8=-4/3+b$; thus $b=\frac{28}{3}$
