Let $(x,y)$ be a curve on $R^2$ defined by the equation $f(x,y)=0$. Suppose $f$ is differentiable and $f_y$ never vanishes. Meanwhile, assume further that all the second-order partial derivatives of $f$ exist and are continuous. Express $d^2y/dx^2$ in terms of $f_x$, $f_y$, $f_{xx}$, $f_{xy}$, $f_{yy}$.
I am definitely sure the main idea in this problem is Implicit Function Theorem to evaluate $dy/dx$, but evaluating second derivative became troublesome for me. In particularly, I could not correlate $d^2y/dx^2$ with $f_{yy}$. Any help would be strongly appreciated!
You have a function $y: \Bbb{R} \to \Bbb{R}$ (or open subsets thereof) such that for all $x$, $f(x,y(x)) = 0$. Part of the conclusion of the implicit function theorem is that the function $y(\cdot)$ will be twice continuously differentiable (because $f$ is). Now, differentiate both sides with respect to $x$: \begin{align} \dfrac{d}{dx} \bigg|_x f(x,y(x)) &= 0 \end{align} Now, by the chain rule (which can be applied since by the implicit function theorem, all the functions involved are atleast $C^2$), \begin{align} \dfrac{\partial f}{\partial x} \bigg|_{(x,y(x))} + \dfrac{\partial f}{\partial x} \bigg|_{(x,y(x))} \cdot y'(x) &= 0 \end{align} Now, since this equation is once again true for all $x \in \Bbb{R}$, we can differentiate again. Use the chain rule on the first term, and product (and chain) rule on second term. If we differentiate the first term, we get: \begin{align} \dfrac{d}{dx} \bigg|_{x} \left( \dfrac{\partial f}{\partial x} \bigg|_{(x,y(x))} \right) &= \dfrac{\partial^2 f}{\partial x^2} \bigg|_{(x,y(x))} + \dfrac{\partial^2 f}{\partial y\partial x} \bigg|_{(x,y(x))} \cdot y'(x). \end{align} I leave it to you to differentiate carefully the second term. You should get a term involving $y''(x)$. Then, simply move everything over to the other side to solve for $y''(x)$ in terms of $y'(x)$, and the various partial derivatives of $f$ (evaluated of course at appropriate points).