The correct answer is: A. continuous but not differentiable.
The function $f(x) = |x|^3$ is continuous at $x=0$ because the left and right hand limits are equal to the function value at $x=0$. However, the function is not differentiable at $x=0$ because the derivative does not exist at $x=0$.
The derivative of a function at a point is the limit of the difference quotient as the difference in the $x$-values approaches zero. In other words, the derivative of a function at a point is the slope of the line tangent to the function at that point.
The function $f(x) = |x|^3$ is not differentiable at $x=0$ because the line tangent to the function at $x=0$ does not exist. This is because the function has a sharp corner at $x=0$.
A function can be continuous but not differentiable at a point. This happens when the function has a sharp corner at that point.
A function can also be differentiable but not continuous at a point. This happens when the function has a jump discontinuity at that point.
A function can be both continuous and differentiable at a point. This is the most common case.