Let $\displaystyle J[y]=\int_{\alpha}^{\beta}f[x,y(x),y'(x)]dx$, where $y(\alpha)=a$, $y(\beta)=b$. Consider a small perturbation $\delta y$. To second order, we have
$$J[y+\delta y]-J[y]=\int_{\alpha}^{\beta}\left\{\delta y\frac{\partial f}{\partial y}+(\delta y)'\frac{\partial f}{\partial y'}\right\}dx+\frac{1}{2}\int_{\alpha}^{\beta}\left\{(\delta y)^2\frac{\partial^2 y}{\partial y^2}+2\delta y(\delta y)'\frac{\partial^2 f}{\partial y\partial y'}+(\delta y)'^2\frac{\partial^2 f}{\partial y'^2}\right\}dx$$
where $\delta y(\alpha)=\delta y(\beta)=0$. Integrating the first term ("first variation" $\delta J[y,\delta y]$) by parts we get
$$\int_{\alpha}^{\beta}\left\{\delta y\frac{\partial f}{\partial y}+(\delta y)'\frac{\partial f}{\partial y'}\right\}dx=\int_{\alpha}^{\beta}\delta y\left\{\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)\right\}dx$$
For the second term, note that $\delta y(\delta y)'=(\delta y^2)'/2$, so, integrating by parts, we can rewrite this "second variation" $\delta^2 J[y,\delta y]$ as
$$\frac{1}{2}\int_{\alpha}^{\beta}\left\{(\delta y)^2\frac{\partial^2 y}{\partial y^2}+2\delta y(\delta y)'\frac{\partial^2 f}{\partial y\partial y'}+(\delta y)'^2\frac{\partial^2 f}{\partial y'^2}\right\}dx=\frac{1}{2}\int_{\alpha}^{\beta}\left\{(\delta y)^2\left[\frac{\partial^2 f}{\partial y^2}-\frac{d}{dx}\left(\frac{\partial^2 f}{\partial y\partial y'}\right)\right]+(\delta y)'^2\frac{\partial^2 f}{\partial y'^2}\right\}dx$$
Hence, to second order
$$J[y+\delta y]-J[y]=\delta J[y,\delta y]+\delta^2 J[y,\delta y]=\int_{\alpha}^{\beta}\delta y\left\{\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)\right\}dx+\frac{1}{2}\int_{\alpha}^{\beta}\left\{(\delta y)^2\left[\frac{\partial^2 f}{\partial y^2}-\frac{d}{dx}\left(\frac{\partial^2 f}{\partial y\partial y'}\right)\right]+(\delta y)'^2\frac{\partial^2 f}{\partial y'^2}\right\}dx$$
Suppose that $\delta^2 J[y,\delta]\geq 0$ for all sufficiently differentiable functions $y(x)$ and $\delta y$ satisfying $\delta y(\alpha)=\delta y(\beta)=0$ and $y(\alpha)=a$, $y(\beta)=b$. Assume that $y_0(x)$ satisfies $\displaystyle \frac{\partial f}{\partial y_0}-\frac{d}{dx}\left(\frac{\partial f}{\partial y_0'}\right)\equiv 0$. How can one show that $J[y]$ has an absolute minimum for $y=y_0$?