How to solve $f\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}=0$? (where $f(x,t)$ is assumed to be in $C^\infty(\mathbb{R}\times\mathbb{R^+}\rightarrow\mathbb{R})$)
I can find a particular solution which is $f=\frac{x}{t}$. Is this the only solution? If not how can I find all the other solution?
Note that if $f(x,t)$ satisfies the given condition, then $f(x-a,t-b)$ also satisfies the given condition.
Along the curve $x(t)$, $$ \frac{\mathrm{d}f}{\mathrm{d}t}=\frac{\mathrm{d}x}{\mathrm{d}t}\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}\tag{1} $$ the given condition implies that $f$ will remain constant on curves where $$ \frac{\mathrm{d}x}{\mathrm{d}t}=f\tag{2} $$ Suppose $f(x_0,t_0)=a$. Then $f(x,t)=a$ on the line where $\frac{x-x_0}{t-t_0}=a$.
Suppose we know that $f(x,0)=\phi(x)$, then $$ f(x+\phi(x)t,t)=\phi(x)\tag{3} $$
If we use $\phi(x)=x$ in $(3)$, we get $$ f(x+xt,t)=x\implies f(x,t)=\frac x{t+1}\tag{4} $$ which is a translate of your function $f(x,t)=\frac xt$.
If we use $\phi(x)=x^2$ in $(3)$, we get $$ f(x+x^2t,t)=x^2\implies f(x,t)=\left(\frac{-1+\sqrt{1+4xt}}{2t}\right)^2\tag{5} $$ Thus, we can generate different functions $f$ given different functions $\phi$.
O.L. mentions that the equation $$ f\frac{\partial f}{\partial x}+\frac{\partial f}{\partial t}=0\tag{6} $$ is called the Inviscid Burgers' Equation.
