We consider $v = \frac{\partial}{\partial x}$ and $w = x * \frac{\partial}{\partial z} + \frac{\partial}{\partial y}$.
I need to first find the Lie bracket between them which i get to be: $\frac{\partial}{\partial x} * x * \frac{\partial}{\partial z} - x * \frac{\partial^{2}}{\partial z \partial x}.$
Now assuming $f$ is smooth, and $L_{v}f = L_{w}f = 0,$ i need to show that $f$ is constant.
I am trying to show that $L_{u}f = 0$ for all vectors $u$ first, and then use $f(\gamma(1)) - f(\gamma(0)) = \int_0^1 L_{\gamma(t)}f\,dt.$
But i am at a loss because, i need the gradient of $f$ to compute $L_{u}f$.
Thanks in advance.
You messed something up in the first computation - the Lie bracket of two vector fields is always a vector field. In this case we have $$[\partial_x, x\partial_z + \partial_y] = [\partial_x, \partial_y] + \partial_x\circ(x\partial_z) - x\partial_z \circ \partial_x = (\partial_x x)\partial_z = \partial_z.$$
For the second question, remember that the Lie derivative of a function is just the usual directional derivative - so if we can show $L_u f = 0$ for all $u$ in a basis/frame then $f$ must be (locally) constant. We are given $L_v f = L_w f = 0$, which implies that $L_{[v,w]} f = (L_v L_w -L_w L_v)f=L_v(0)-L_w(0)=0$. We now have the facts
$$ \partial_x f = x \partial_z f + \partial_y f = \partial_z f = 0.$$
Substituting $\partial_z f = 0$ into the middle expression gives $\partial_y f = 0$; so we have $\partial_x f = \partial_y f = \partial_z f = 0$.