The Euler–Lagrange equation

Fix an arbitrary smooth function $\eta(x)$ with $\eta(x_0)=\eta(x_1)=0$, and consider the one-parameter family of competitor curves $y_\epsilon(x) = y(x) + \epsilon\,\eta(x)$, all of which share the same endpoint values as $y$. Define $\phi(\epsilon) = J[y_\epsilon]$; since $y$ is assumed to extremize $J$, the ordinary function $\phi$ has a critical point at $\epsilon=0$, so $\phi'(0)=0$.

Differentiating under the integral sign, $\phi'(\epsilon) = \int_{x_0}^{x_1} \left[ \frac{\partial L}{\partial y} \,\eta + \frac{\partial L}{\partial y'} \,\eta' \right] dx$, where the partial derivatives of $L$ are evaluated along $y_\epsilon$. Setting $\epsilon=0$ gives $\int_{x_0}^{x_1} \left[ \frac{\partial L}{\partial y} \,\eta + \frac{\partial L}{\partial y'} \,\eta' \right] dx = 0$.

Integrate the second term by parts: $\int_{x_0}^{x_1} \frac{\partial L}{\partial y'} \,\eta'\,dx = \left[ \frac{\partial L}{\partial y'} \,\eta \right]_{x_0}^{x_1} - \int_{x_0}^{x_1} \frac{d}{dx}\frac{\partial L}{\partial y'} \,\eta\,dx$. The boundary term vanishes because $\eta(x_0)=\eta(x_1)=0$, leaving $\int_{x_0}^{x_1} \left[ \frac{\partial L}{\partial y} - \frac{d}{dx}\frac{\partial L}{\partial y'} \right] \eta\,dx = 0$.

This integral vanishes for every admissible $\eta$. By the fundamental lemma of the calculus of variations — if a continuous function integrates to zero against every such test function, the function itself must be identically zero — the bracketed quantity vanishes at every $x$, which is exactly $\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y'}=0$.