Area under curves functional analysis question Consider the functions defined inplicitly by te equation $y^{3}-3y+x=0$ on various intervals in the real line . If $x\in(-\infty,2)\cup(2,\infty)$ the equation implicitly defines a unique real valued differentiable function $y=f(x)$ . if $x\in (-2,2)$, the equation implicitly defines a unique real valued differentiable function $y=g(x)$ satisfying $g(0)=0$. 
Q1) If $f(-10\sqrt{2}) =2\sqrt{2}$ , then $f''(-10\sqrt{2})$ = ?
Q2) $\int^{1}_{-1}g'(x)dx$ =
Q3) The area of the region bounded by the curve $y=f(x)$, the x-axis and the lines $x=a$ and $x=b$ where $-\infty<a<b<-2$, ?

$(1)\;\;$ Given $$y^3-3y+x=0\;,...........(1)$$ Now Differentiate both side w. r to $x\;,$ We get
$$\displaystyle 3y^2\cdot y'-3y'+1=0....................(2)\Rightarrow y'=\frac{1}{3-3y^2}\;,$$ Where $$\displaystyle y=f(x)\;\;, y'=\frac{dy}{dx}=f'(x)\;\;,y''=f''(x)=\frac{d^2y}{dx^2}$$
Now Given $\displaystyle y=f\left(-10\sqrt{2}\right)=2\sqrt{2}.$
So we get $$\displaystyle y' = \frac{1}{3-3y^2}=\frac{1}{3-3(2\sqrt{2})^2} = -\frac{1}{21}$$
Now Again Differentiate $(2)\;,$ We get
$$\displaystyle 3y^2\cdot y''+y'\cdot 6yy'-3y''=0\Rightarrow 6y(y')^2+3y^2y''-3y''=0$$
Now put $\displaystyle y = 2\sqrt{2}$ and $\displaystyle y'=-\frac{1}{21}\;,$ We get $\displaystyle y''= \frac{2y(y')^2}{1-3y^2}=-\frac{6\cdot 2\sqrt{2}}{(21)^3} = -\frac{4\sqrt{2}}{243\times9}$ 
So we get $\displaystyle f''\left(-10\sqrt{2}\right) = -\frac{4\sqrt{2}}{243\times 9}$
$(2)\;\;$ We have $$\displaystyle g'(x) = \frac{dy}{dx} = \frac{1}{3[1-(f(x))^2]}$$ (Even function)
So $$\displaystyle \int_{-1}^{1} g'(x) = 2\int_{0}^{1}g'(x)dx = 2\left[g(x)\right]_{0}^{1} = 2[g(1)-g(0)] = 2g(0)\;\;,$$ bcz given $g(0)=0$
$(3)\;\;$ The Required Area $$\displaystyle = \int_{a}^{b}f(x)\cdot 1 dx = \left[x\cdot f(x)\right]_{a}^{b}-\int_{a}^{b}xf'(x)dx$$
$$\displaystyle =bf(b)-af(a)+\int_{a}^{b}\frac{x}{3\left[(f(x))^2-1\right]}dx$$
Bcz from equation no. $(2)\;,$ We have $\displaystyle y'=\frac{dy}{dx}=\frac{1}{3-3y^2}$
