Te definition of the functional derivative of a functional $I[g]$ is the distribution $\frac{\delta I}{\delta g}(\tau)$ such that $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle := \frac{d}{d\alpha}|_{\alpha=0} I[g+ \alpha h]$$ for every test function $h$. In our case, assuming to deal with functions which suitably vanish before reaching $\pm \infty$, $$I[g] = \int_{-\infty}^t g(x)dx$$ so that $$I[\dot{f}]= f(t)$$ as requested. Going on with the procedure $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle = \frac{d}{d\alpha}|_{\alpha=0} \int_{-\infty}^t(g(\tau)+ \alpha h(\tau)) d\tau = \int_{-\infty}^t h(\tau) dx = \int_{-\infty}^{+\infty} \theta(t-\tau)h(\tau) d\tau$$ where $\theta(\tau)=1$ for $\tau\geq 0$ and $\theta(\tau)=0$ for $\tau<0$ and so $$\frac{\delta f(t)}{\delta \dot{f}}(\tau) = \frac{\delta I}{\delta g}(\tau)= \theta(t-\tau)$$

The definition of the functional derivative of a functional $I[g]$ is the distribution $\frac{\delta I}{\delta g}(\tau)$ such that $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle := \frac{d}{d\alpha}\bigg\rvert_{\alpha=0} I[g+ \alpha h]$$ for every test function $h$. In our case, assuming to deal with functions which suitably vanish before reaching $\pm \infty$, $$I[g] = \int_{-\infty}^t g(x)dx$$ so that $$I[\dot{f}]= f(t)$$ as requested. Going on with the procedure $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle = \frac{d}{d\alpha}|_{\alpha=0} \int_{-\infty}^t(g(\tau)+ \alpha h(\tau)) d\tau = \int_{-\infty}^t h(\tau) dx = \int_{-\infty}^{+\infty} \theta(t-\tau)h(\tau) d\tau$$ where $\theta(\tau)=1$ for $\tau\geq 0$ and $\theta(\tau)=0$ for $\tau<0$ and so $$\frac{\delta f(t)}{\delta \dot{f}}(\tau) = \frac{\delta I}{\delta g}(\tau)= \theta(t-\tau)$$

Te definition of the functional derivative of a functional $I[g]$ is the distribution $\frac{\delta I}{\delta g}(\tau)$ such that $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle := \frac{d}{d\alpha}|_{\alpha=0} I[g+ \alpha h]$$ for every test function $h$. In our case, assuming to deal with functions which suitably vanish before reaching $\pm \infty$, $$I[g] = \int_{-\infty}^t g(x)dx$$ so that $$I[\dot{f}]= f(t)$$ as requested. Going on with the procedure $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle = \frac{d}{d\alpha}|_{\alpha=0} \int_{-\infty}^t(g(\tau)+ \alpha h(\tau)) d\tau = \int_{-\infty}^t h(\tau) dx = \int_{-\infty}^{+\infty} \theta(t-\tau)h(\tau) d\tau$$ where $\theta(\tau)=1$ for $\tau\geq 0$ and $\theta(\tau)=0$ for $\tau<0$ and so $$\frac{\delta f(t)}{\delta \dot{f(t)}} = \frac{\delta I}{\delta g}(\tau)= \theta(t-\tau)$$

Te definition of the functional derivative of a functional $I[g]$ is the distribution $\frac{\delta I}{\delta g}(\tau)$ such that $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle := \frac{d}{d\alpha}|_{\alpha=0} I[g+ \alpha h]$$ for every test function $h$. In our case, assuming to deal with functions which suitably vanish before reaching $\pm \infty$, $$I[g] = \int_{-\infty}^t g(x)dx$$ so that $$I[\dot{f}]= f(t)$$ as requested. Going on with the procedure $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle = \frac{d}{d\alpha}|_{\alpha=0} \int_{-\infty}^t(g(\tau)+ \alpha h(\tau)) d\tau = \int_{-\infty}^t h(\tau) dx = \int_{-\infty}^{+\infty} \theta(t-\tau)h(\tau) d\tau$$ where $\theta(\tau)=1$ for $\tau\geq 0$ and $\theta(\tau)=0$ for $\tau<0$ and so $$\frac{\delta f(t)}{\delta \dot{f}}(\tau) = \frac{\delta I}{\delta g}(\tau)= \theta(t-\tau)$$