How do you get the second order implicit differentiation? We were given an exercise in school:
Given $x^2 + 25y^2 = 100$, show that 
$$\frac{dy}{dx^2} = -\frac{4}{25y^3}$$
I am stuck on the first order which is 
$$-\frac{x}{25y}$$
When I'm now going to the second, I always end up with the wrong answer where the denominator is somewhere at $625y^2$. I know that you have to use the quotient rule on the first order derivative to get to the second, but I don't know how. Can anyone answer with a detailed step-by-step solution? I'm really lost. Our prof didn't explain that topic well to us.
 A: hint: $(x^2+25y^2)' = 0 \to 2x+50yy' = 0 \to y' = -\dfrac{x}{25y} \to y'' = \left(\dfrac{-x}{25y}\right)' = \dfrac{....}{625y^2}$
A: You were given 
$$x^2 + 25y^2 = 100\tag{1}$$
from which you correctly determined that 
$$y' = -\frac{x}{25y}\tag{2}$$
We will need both of these equations later.
If we differentiate 
$$y' = -\frac{x}{25y}$$
implicitly with respect to $x$, we obtain
\begin{align*}
y'' & = -\frac{25y - 25xy'}{625y^2} && \text{by the Quotient Rule}\\
    & = -\frac{y - xy'}{25y^2} && \text{cancel a factor of $25$}\\
    & = -\frac{y - x\left(-\dfrac{x}{25y}\right)}{25y^2} && \text{use equation (2) to substitute $-\dfrac{x}{25y}$ for $y'$}\\
    & = -\frac{\dfrac{25y^2 + x^2}{25y}}{25y^2} && \text{add fractions in numerator}\\
    & = -\frac{25y^2 + x^2}{625y^3} && \text{multiply numerator and denominator by $25y$}\\
   & = -\frac{100}{625y^3} && \text{use equation (1) to substitute $100$ for $x^2 + 25y^2$}\\
   & = -\frac{4}{25y^3} && \text{cancel a factor of $25$}
\end{align*}
