Step 1 :The function \(f(x)=\sqrt[3]{x}\) is defined for all real numbers, including \(x=-8\). Therefore, option a is incorrect.
Step 2 :The limit as \(x\) approaches \(-8\) also exists, so option c is incorrect.
Step 3 :We need to check if the derivative of the function at \(x=-8\) exists to determine if the function is differentiable at that point.
Step 4 :The derivative of \(f(x)=\sqrt[3]{x}\) is \(f'(x)=\frac{1}{3\sqrt[3]{x^2}}\).
Step 5 :We need to check if this derivative is defined at \(x=-8\).
Step 6 :The derivative of the function at \(x=-8\) is defined and is equal to \(-\frac{1}{12}\).
Step 7 :Therefore, the function \(f(x)=\sqrt[3]{x}\) is differentiable at \(x=-8\).
Step 8 :Final Answer: \(\boxed{\text{(d) } f(x)=\sqrt[3]{x} \text{ is differentiable at } x=-8}\)