Curvature of implicit curve This is a problem from "Advanced Calculus" by Buck:
If $\gamma$ is a curve which satisfies the equation $f(x,y)=0$, show that the curvature at a point on $\gamma$ is given by:
$$\kappa = \frac{|f_{xx} f_y^2-2f_{xy} f_x f_y + f_x^2 f_{yy}|}{|f_x^2+f_y^2|^{3/2}}$$
Sincerely, I consider that the textbook does not provide enough theory to tackle problems of this nature. Can you give me a suggestion or at least a good reference?
 A: Coincidentally, I just solved this problem a month or two ago (in a different context), so I'll take my notes from there and reconstruct my solution.
Starting with:
$$f(x,y)=0$$
Take the derivative of both sides, with respect to x:
$$\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}=0$$
(This is because $\frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}$, for any $f(x,y)$.) Solve for $\frac{dy}{dx}$:
$$\frac{dy}{dx}=-{\frac{\partial f}{\partial x}\over\frac{\partial f}{\partial y}}$$
Taking the derivative of both sides again, we get:
$$\frac{d^2y}{dx^2}=-{\frac{\partial f}{\partial y}\frac d{dx}\left(\frac{\partial f}{\partial x}\right)-\frac{\partial f}{\partial x}\frac d{dx}\left(\frac{\partial f}{\partial y}\right)\over\left(\frac{\partial f}{\partial y}\right)^2}$$
But, since $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are still functions of $x$ and $y$, we can use the above rule:
$$\frac{d^2y}{dx^2}=-{\frac{\partial f}{\partial y}\left(\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial x\partial y}\frac{dy}{dx}\right)-\frac{\partial f}{\partial x}\left(\frac{\partial^2f}{\partial x\partial y}+\frac{\partial^2f}{\partial y^2}\frac{dy}{dx}\right)\over\left(\frac{\partial f}{\partial y}\right)^2}$$
And since we already solved for $\frac{dy}{dx}$, we can substitute that in, too:
$$\frac{d^2y}{dx^2}=-{\frac{\partial f}{\partial y}\left(\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial x\partial y}\left(-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial y}\right)\right)-\frac{\partial f}{\partial x}\left(\frac{\partial^2f}{\partial x\partial y}+\frac{\partial^2f}{\partial y^2}\left(-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial y}\right)\right)\over\left(\frac{\partial f}{\partial y}\right)^2}$$
which simplifies to:
$$\frac{d^2y}{dx^2}=-{\frac{\partial^2f}{\partial x^2}\left(\frac{\partial f}{\partial y}\right)^2-2\frac{\partial^2f}{\partial x\partial y}\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}+\frac{\partial^2f}{\partial y^2}\left(\frac{\partial f}{\partial x}\right)^2\over\left(\frac{\partial f}{\partial y}\right)^3}$$
Now, we can plug them into the curvature formula (which I already derived here):
$$\kappa={\frac{dx}{dt}\frac{d^2y}{dt^2}-\frac{dy}{dt}\frac{d^2x}{dt^2}\over\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}^3}$$
If we let $t=x$ (essentially turning the parametric function input into an implicit or explicit function input), we get:
$$\kappa={\frac{dx}{dx}\frac{d^2y}{dx^2}-\frac{dy}{dx}\frac{d^2x}{dx^2}\over\sqrt{\left(\frac{dx}{dx}\right)^2+\left(\frac{dy}{dx}\right)^2}^3}$$
which simplifies to:
$$\kappa={\frac{d^2y}{dx^2}\over\sqrt{1+\left(\frac{dy}{dx}\right)^2}^3}$$
Then we substitute:
$$\kappa={-({\frac{\partial^2f}{\partial x^2}\left(\frac{\partial f}{\partial y}\right)^2-2\frac{\partial^2f}{\partial x\partial y}\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}+\frac{\partial^2f}{\partial y^2}\left(\frac{\partial f}{\partial x}\right)^2)/\left(\frac{\partial f}{\partial y}\right)^3}\over\sqrt{1+\left(-\frac{\partial f}{\partial x}/\frac{\partial f}{\partial y}\right)^2}^3}$$
which simplifies to:
$$\kappa=-{\frac{\partial^2f}{\partial x^2}\left(\frac{\partial f}{\partial y}\right)^2-2\frac{\partial^2f}{\partial x\partial y}\frac{\partial f}{\partial x}\frac{\partial f}{\partial y}+\frac{\partial^2f}{\partial y^2}\left(\frac{\partial f}{\partial x}\right)^2\over\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}^3}$$
as desired (within a plus or minus sign). [For whatever reason, I couldn't get those parentheses in the second-to-last expression to work properly.]
(If $\frac{\partial f}{\partial y}=0$, then swap $x$ and $y$. If both are zero, then $\kappa$ is undefined to begin with, without the use of L'Hôpital's rule.)
Originally, I actually did this another way; if you like, I can post that way as well.
A: We can assume that the curve is smooth in a neighbourhood of the point that we are considering. If it weren't, then curvature would not make sense. 
Our curve is given by an equation $f(x,y)=0$. 
Let us assume that a curve $\gamma(t) = (x(t),y(t))$ parametrises our curve. This means what
$$f(x(t),y(t)) \equiv 0$$
The formula for the curvature of a plane curve is:
$$\kappa = \frac{\dot x \ddot y - \ddot x \dot y}{(\dot x^2 + \dot y^2)^{3/2}}$$
If we can express $\dot x$, $\dot y$, $\ddot x$ and $\ddot y$ in terms of $f_x$, $f_y$, $f_{xx}$, $f_{xy}$ and $f_{yy}$ then we're done.
Differentiating the identity $f(x(t),y(t)) \equiv 0$ with respect to $t$ gives $\dot x f_x + \dot y f_y \equiv 0$. It follows that, for some constant $\lambda > 0$, we have $\dot x \equiv -\lambda f_y$ and $\dot y \equiv \lambda f_x$. 
Differentiating $\dot x \equiv -\lambda f_y$ with respect to $t$ gives $\ddot x \equiv -\lambda(\dot x f_{xy} + \dot y f_{yy})$, while differentiating $\dot y \equiv \lambda f_x$ with respect to $t$ gives $\ddot y \equiv \lambda(\dot x f_{xx} + \dot y f_{xy})$. 
We can now put these into the expression for $\kappa$. Firstly:
\begin{eqnarray*}
\dot x \ddot y - \ddot x \dot y &\equiv& -\lambda^2 f_y(\dot x f_{xx} + \dot y f_{xy})+\lambda^2(\dot x f_{xy} + \dot y f_{yy})f_x \\ \\
&\equiv& -\lambda^2 f_y(-\lambda f_y f_{xx} + \lambda f_x f_{xy})+\lambda^2(-\lambda f_y f_{xy} + \lambda f_x f_{yy})f_x \\ \\
&\equiv& \lambda^3 \left( f_y^2f_{xx}-2f_xf_yf_{xy}+f_x^2f_{yy}\right)
\end{eqnarray*}
Similarly, with the denominator:
\begin{eqnarray*}
\dot x^2 + \dot y^2 &\equiv& (-\lambda f_y)^2 + (\lambda f_x)^2 \\ \\
&\equiv& \lambda^2(f_x^2 + f_y^2) \\ \\
\left(\dot x^2 + \dot y^2\right)^{3/2} &\equiv& \lambda^3 \left(f_x^2 + f_y^2\right)^{3/2}
\end{eqnarray*}
Putting the numerator over the nominator gives the result
$$\frac{\dot x \ddot y - \ddot x \dot y}{(\dot x^2 + \dot y)^{3/2}} \equiv 
\frac{f_y^2f_{xx}-2f_xf_yf_{xy}+f_x^2f_{yy}}{\left(f_x^2 + f_y^2\right)^{3/2}}$$
The modulus signs can be added. Some authors insist that $\kappa \ge 0$, while others allow $\kappa \in \mathbb R$.
