1. Problem statement: Find the stationary point(s) of the curve $y=(3x^2+8)^{5/3}$ and determine their nature. 2. Formula and rules used: We use the chain rule: if $y=u^n$ then $y'=n u^{n-1}u'$. Important rule: $3x^2+8>0$ for all real $x$, so $(3x^2+8)^{2/3}>0$ for all real $x$ and fractional powers are real here. 3. Compute the derivative (showing intermediate algebra): Start with $y=(3x^2+8)^{5/3}$. Differentiate using the chain rule: $y'=\frac{5}{3}(3x^2+8)^{2/3}\cdot(6x)$. Simplify the numeric factor with cancellation: $\frac{5}{3}\cdot 6 = 5\cdot\frac{\cancel{6}}{\cancel{3}} = 5\cdot 2 = 10$. Therefore $y'=10x(3x^2+8)^{2/3}$. 4. Solve $y'=0$ to find stationary points: Set $10x(3x^2+8)^{2/3}=0$. Divide both sides by 10 showing cancellation: $\frac{\cancel{10}x(3x^2+8)^{2/3}}{\cancel{10}}=0$. So $x(3x^2+8)^{2/3}=0$. Because $(3x^2+8)^{2/3}>0$ for every real $x$, the only solution is $x=0$. Thus the stationary point abscissa is $x=0$. 5. Find the corresponding $y$ value: Evaluate $y(0)=(3\cdot 0^2+8)^{5/3}=8^{5/3}$. Write $8=2^3$ and simplify: $8^{5/3}=(2^3)^{5/3}=2^5=32$. Therefore the stationary point is $(0,32)$. 6. Determine the nature via the second derivative test: Start from $y'=10x(3x^2+8)^{2/3}$ and apply the product rule. Compute $y''=10(3x^2+8)^{2/3}+10x\cdot\frac{2}{3}(3x^2+8)^{-1/3}\cdot(6x)$. Simplify the factor $\frac{2}{3}\cdot 6x$ with cancellation: $\frac{2}{3}\cdot 6x = 2\cdot\frac{\cancel{6}x}{\cancel{3}} = 4x$. So $y''=10(3x^2+8)^{2/3}+10x\cdot 4x(3x^2+8)^{-1/3} = 10(3x^2+8)^{2/3}+40x^2(3x^2+8)^{-1/3}$. Factor out $(3x^2+8)^{-1/3}$: $y''=(3x^2+8)^{-1/3}\bigl(10(3x^2+8)+40x^2\bigr)$. Simplify inside: $10(3x^2+8)+40x^2=30x^2+80+40x^2=70x^2+80$. Thus $y''=(3x^2+8)^{-1/3}(70x^2+80)$. Evaluate at $x=0$: $y''(0)=8^{-1/3}\cdot 80$. Since $8^{1/3}=2$, $y''(0)=80/2=40>0$. A positive second derivative means the stationary point is a local minimum. 7. Final answer (concise): The curve has a single stationary point at $(0,32)$, and it is a local minimum.