Step-by-step Solution:

1. Understand the Goal: We are asked to evaluate the definite integral $I = \int\limits_{ - 1}^1 {g'\left( x \right)dx}$ where $y=g(x)$ is an implicitly defined function.

2. Apply the Fundamental Theorem of Calculus: The Fundamental Theorem of Calculus states that if $G'(x) = g(x)$, then $\int_a^b g(x) dx = G(b) - G(a)$. In our case, the integrand is $g'(x)$, and its antiderivative is $g(x)$. Therefore, we can directly evaluate the integral as: $I = \int\limits_{ - 1}^1 {g'\left( x \right)dx} = \left[ {g\left( x \right)} \right]_{ - 1}^1 = g\left( 1 \right) - g\left( { - 1} \right)$

3. Analyze the Properties of the function $g(x)$: The function $y = g(x)$ is defined by the equation $y^3 - 3y + x = 0$ for $x \in (-2, 2)$ with the condition $g(0) = 0$. Let's check for symmetry in the defining equation. Let's replace $(x, y)$ with $(-x, -y)$ in the equation: $(-y)^3 - 3(-y) + (-x) = 0$ $-y^3 + 3y - x = 0$ $-(y^3 - 3y + x) = 0$ This is the same as the original equation. This means that if a point $(x, y)$ lies on the curve, then the point $(-x, -y)$ also lies on the curve.

4. Prove that $g(x)$ is an odd function: We are given that for $x \in (-2, 2)$, there is a unique differentiable function $y = g(x)$ that satisfies $y^3 - 3y + x = 0$ and $g(0) = 0$. Let's define a new function $h(x) = -g(-x)$. We want to check if $h(x)$ satisfies the same conditions as $g(x)$.

   • Check the implicit equation: Substitute $y = h(x)$ into the equation: $(h(x))^3 - 3(h(x)) + x = (-g(-x))^3 - 3(-g(-x)) + x$ $= -[g(-x)]^3 + 3g(-x) + x$ Since $g(x)$ satisfies the original equation, for any input, say $u$, we have $(g(u))^3 - 3g(u) + u = 0$. Let's set $u = -x$: $(g(-x))^3 - 3g(-x) + (-x) = 0 \implies [g(-x)]^3 - 3g(-x) = x$ Now substitute this back into our expression for $h(x)$: $= -(x) + x = 0$ So, $h(x)$ also satisfies the implicit equation $y^3 - 3y + x = 0$.

   • Check the initial condition: $h(0) = -g(-0) = -g(0) = -0 = 0$ So, $h(0) = 0$.

   Since both $g(x)$ and $h(x) = -g(-x)$ satisfy the same defining equation and the same initial condition, and the problem states that such a function is unique, we must have: $g(x) = h(x) \implies g(x) = -g(-x)$ This shows that $g(x)$ is an odd function.

5. Use the Odd Function Property to Finalize the Calculation: Since $g(x)$ is an odd function, we have the property $g(-x) = -g(x)$. For $x=1$, this gives: $g(-1) = -g(1)$ Now substitute this back into our expression for the integral from Step 2: $I = g(1) - g(-1) = g(1) - (-g(1)) = g(1) + g(1) = 2g(1)$

6. Conclusion: The value of the integral is $2g(1)$. Comparing this with the given options: A: $2g(-1) = 2(-g(1)) = -2g(1)$ B: $0$ C: $-2g(1)$ D: $2g(1)$

   The correct option is D.