I'm stuck in finding the general solution for this DE because I'm not sure what technique I should use:
$$\frac{dv}{dx}=\frac{g}{v}-\frac{k}{m}$$
Where g, k and m are constants.
I'm not sure if I can really use substitution method here. I've also tried long division but I always end up having a higher degree of the numerator. Partial fractions also doesn't seem to work as the denominator is not factorable when considered as a separable equation.
This equation is separable. You write:
$$dv = \left(\frac{g}{v} -\frac{k}{m}\right)\cdot dx $$ $$ \Rightarrow \frac{dv}{\frac{g}{v} -\frac{k}{m}} = dx$$ $$ \Rightarrow \int_{v_0}^{v} \frac{dv}{\frac{g}{v} -\frac{k}{m}} = \int_0^xdx$$ $$\Rightarrow \int_{v_0}^{v} \frac{v}{g-\frac{k}{m}v} dv = x $$
Now you have to look up this integral on some integral table and solve for $x$.
The answer on your question is given by @JakobElias. But I want to give some extra information.
When we have an oject falling vertically and we have drag we can use the following system of equations:
$$ \begin{cases} \text{g}=\frac{\text{d}^2\space\text{h}\left(t\right)}{\text{d}\space t^2}+\frac{\rho\cdot\text{A}\cdot\text{C}_\text{d}}{2\cdot\text{m}}\cdot\left(\frac{\text{d}\space\text{h}\left(t\right)}{\text{d}\space t}\right)^2\\ \\ \text{V}_\text{T}=\sqrt{\frac{2\cdot\text{m}\cdot\text{g}}{\rho\cdot\text{A}\cdot\text{C}_\text{d}}}\\ \\ \text{C}_\text{d}=\frac{\text{F}_\text{d}}{\frac{1}{2}\cdot\rho_\text{m}\cdot\text{u}^2\cdot\text{A}_\text{m}} \end{cases}\tag1 $$
Where $\text{h}$ is the fall distance, $\rho$ is the fluid density, $\text{m}$ is the mass, $\text{C}_\text{d}$ is the drag coefficient, $\text{A}$ is the projected surface area, $\text{g}$ is the gravitational acceleration, $t$ is time, $\text{V}_\text{T}$ is the terminal velocity, $\text{F}_\text{d}$ is the drag force, $\rho_\text{m}$ is the mass density, $\text{u}$ is the characteristic speed and $\text{A}_\text{m}$ is the area.
The correct equation of motion for a free-falling object as written in the OP is incorrect. Rather, we have
$$\frac{dv}{dt}=mg-kv^2 \tag 1$$
Dividing by $v=\frac{dx}{dt}$ reveals that
$$\frac{dv}{dx}=\frac{g}{v}-\frac{k}m v$$
Then, we have
$$\int_{x_0}^x dx' = \frac{m}{k}\int_{v_0}^v \frac{\nu}{\frac{mg}k-\nu^2}\,d\nu \tag 2$$
where we note that the integration limits cannot embed $v=\sqrt{mg/k}=v_{\infty}$, which is the terminal velocity. Aside, in $(2)$, $v_0$ is the speed of the object when $x=x_0$.
The solution to $(2)$ is given by
$$x=x_0+\frac{m}{2k}\log\left|\frac{v_{\infty}^2-v_0^2}{v_{\infty}^2-v^2}\right| \tag 3$$
Unfortunately, $(3)$ does not provide much insight into the dynamics of the free-falling object. It provides only a relationship between the position of the object and its speed at that position.
It is better to rearrange $(1)$ as
$$\int_{0}^t d\tau =\frac{m}{k}\int_{v(0)}^{v(t)} \frac{1}{v_{\infty}^2-\nu^2}\,d\nu \tag 4$$
We can evaluate the integral on the right-hand side of $(4)$ by enforcing the substitution $\nu = v_{\infty}\tanh(z)$. Proceeding we find that
$$t=\sqrt{\frac{m}{kg}} \left(\text{artanh}(v(t)/v_{\infty})-\text{artanh}(v(0)/v_{\infty})\right)\tag 5$$
Solving $(5)$ for $v(t)$ reveals
$$v(t)=v_{\infty}\left(\frac{\frac{1+v(0)/v_{\infty}}{1-v(0)/v_{\infty}}e^{2t/\sqrt{mg/k}}-1}{\frac{1+v(0)/v_{\infty}}{1-v(0)/v_{\infty}}e^{2t/\sqrt{mg/k}}+1}\right)$$
which provides a more useful form than $(3)$. For example, we see from $(5)$ that if $v(0)>v_{\infty}$, then $v(t)$ decreases to $v_{\infty}$ while if $v(0)