Inviscid Burgers' equation (Verify the solution) From the inviscid Burgers' equation on $\mathbb{R} \times [0,\infty),$ $$u_t + uu_x = 0,$$ for any differentiable function $f$, $v(x,t) = f(x-ut)$ is an implicit solution to the equation. So I want to verify that $v_t + vv_x = 0.$ Since $$\frac{\partial}{\partial t}v(x,t) = \frac{\partial}{\partial t}f(x-ut) = f'(x-ut) (0 - (u+tu_t)) = -(u(x,t) + tu_t(x,t))f'(x-ut)$$ and $$\frac{\partial}{\partial x}v(x,t) = \frac{\partial}{\partial x}f(x-ut) = f'(x-ut)(1-tu_x) = f'(x-ut)(1 - tu_x(x,t)),$$ $$v_t + vv_x = -(u+tu_t)f'(x-ut) + f(x-ut)(1-tu_x)f'(x-ut).$$ I do not see why $v(x,t)$ is the solution to the equation ?

I think there may be a typo in the problem you're being asked to solve. An implicit solution is $u(x, t) = f(x-ut)$. To see why that's the case, take derivatives, as you did:

$$u_t = -f'(x-ut)(u+tu_t) \implies u_t = -\frac{u f'(x-ut)}{1+tf'(u-xt)},$$ $$u_x = f'(x-ut)(1-tu_x) \implies u_x = \frac{f'(x-ut)}{1+tf'(u-xt)}.$$

So we have $u_t + uu_x = 0$, as required.