The function $v(x,t)$ satisfies: $$\frac{\partial v}{\partial t} = \frac{\partial^2v}{\partial x^2} + \left(\frac{\partial v}{\partial x}\right)^2$$ for $0<x<1$, the initial condition $v(x,0) = 0$ for $0\le x\le1$, and the boundary conditions $$\frac{\partial v}{\partial x} = 1\,\,\,\text{ at }\,\,\,x=0,\,t>0\,\,\,\text{ and }\,\,\,v=0\,\,\,\text{ at }\,\,\,x=1,\,t>0$$ By change of dependent variable defined by $v=\ln u$, $u\ne0$, transforms the problem to: $$\frac{\partial u}{\partial t} = \frac{\partial^2u}{\partial x^2}$$ for $0<x<1$ where $u(x,t)$ satisfies the conditions $u(x,0) = 1$ for $0\le x\le1$ and $$\frac{\partial u}{\partial x} = u\,\,\,\text{ at }\,\,\,x=0,\,t>0\,\,\,\text{ and }\,\,\,u=1\,\,\,\text{ at }\,\,\,x=1,\,t>0$$
There is a lot of information here and I have always struggled with change of variable so I just want suggestions as to what a first step would be.
I take it that our OP intended to write
$\dfrac{\partial v}{\partial t} = \dfrac{\partial v^2}{\partial x^2} + \left ( \dfrac{\partial v}{\partial x} \right )^2 \tag 0$
as the first equation in his/her post.
I walk through the steps of the transformation like this:
with
$v = \ln u, \tag 1$
we have
$u = e^v; \tag 2$
$u_t = v_t e^v, \tag 3$
$u_x = v_x e^v, \tag 4$
$u_{xx} = v_{xx} e^v + v_x v_x e^v = v_{xx}e^v + v_x^2 e_v; \tag 5$
then if
$v_t = v_{xx} + v_x^2, \tag 6$
we find
$v_t e^v = v_{xx}e^v + v_x^2 e^v; \tag 7$
now via (3) and (5) this may be written
$u_t = u_{xx}. \tag 8$
These steps may in fact be readily reversed; if we start with (8) and assume (1), (0) emerges as a necessary consequence.