By the invariance of the Lorentz interval and the fact that the speed of light is the same in both frames we have \begin{align*} -c^2 dt^2 + dx^2 = -c^2 dt'^2 + dx'^2 \end{align*} By considering the set of points with t = const we obtain \begin{align*} dx^2 = -c^2 dt'^2 + dx'^2 \end{align*} By definition of the differential we have \begin{align*} dx &= \frac{\partial x}{\partial t'} dt' + \frac{\partial x}{\partial x'} dx' \\ dx' &= \frac{\partial x}{\partial t} dt + \frac{\partial x}{\partial x} dx = \frac{\partial x}{\partial x} dx \\ dt &= \frac{\partial t}{\partial t'} dt' + \frac{\partial t}{\partial x'} dx' \\ dt' &= \frac{\partial t'}{\partial t} dt + \frac{\partial t'}{\partial x} dx = \frac{\partial t'}{\partial x} dx \end{align*} Substituting the differentials we obtain \begin{align*} dx^2 = - c^2 \left( \frac{\partial t'}{\partial x} dx\right)^2 + \left( \frac{\partial x'}{\partial x} dx \right)^2 \\ dx^2 = - c^2 \left( \frac{\partial t'}{\partial x} \right)^2 dx^2 + \left( \frac{\partial x'}{\partial x} \right)^2 dx^2 \\ dx^2 + c^2 \left( \frac{\partial t'}{\partial x} \right)^2 dx^2 - \left( \frac{\partial x'}{\partial x} \right)^2 dx^2 = 0 \\ dx^2 \left(1 + c^2 \left( \frac{\partial t'}{\partial x} \right)^2 - \left( \frac{\partial x'}{\partial x} \right)^2 \right) = 0 \\ \left(1 + c^2 \left( \frac{\partial t'}{\partial x} \right)^2 - \left( \frac{\partial x'}{\partial x} \right)^2 \right) = 0 \end{align*} Now considering the set of points with x = const we obtain \begin{align*} -c^2 dt^2 = -c^2 dt'^2 + dx'^2 \end{align*} Now substituting the differentials we obtain \begin{align*} -c^2 dt^2 = -c^2 \left( \frac{\partial t'}{\partial t} dt\right)^2 + \left( \frac{\partial x'}{\partial t} dt \right)^2 \\ -c^2 dt^2 = -c^2 \left( \frac{\partial t'}{\partial t}\right)^2 dt^2 + \left( \frac{\partial x'}{\partial t} \right)^2 dt^2 \\ -c^2 dt^2 + c^2 \left( \frac{\partial t'}{\partial t}\right)^2 dt^2 - \left( \frac{\partial x'}{\partial t} \right)^2 dt^2 = 0 \\ dt^2 \left(-c^2 + c^2 \left( \frac{\partial t'}{\partial t}\right)^2 - \left( \frac{\partial x'}{\partial t} \right)^2 \right) = 0 \\ \left(-c^2 + c^2 \left( \frac{\partial t'}{\partial t}\right)^2 - \left( \frac{\partial x'}{\partial t} \right)^2 \right) = 0 \end{align*}
Where do I go from here? If I want to obtain:
\begin{align*} \begin{cases} t' = \gamma \left( t - \frac{v}{c^2} x \right) \\ x' = \gamma \left( x - vt \right) \end{cases} \qquad \text{ where } \gamma \equiv \frac{1}{\sqrt{1 - (v/c)^2}} \ge 1 \end{align*}