Let $u = u(x, y)$ be a known solution of the minimal surface equation $$(1 + u_y^2) u_{xx} + (1+u_x^2) u_{yy} - 2 u_x u_y u_{xy} = 0$$
Now, say $$v(x, y) = \frac{1}{\lambda} u(\lambda x, \lambda y)$$
We then compute the required partial derivatives.
$$v_x = \frac{\partial}{\partial x} \left[ \lambda^{-1} u(\lambda x, \lambda y) \right] = \frac{1}{\lambda} \left[ \lambda \cdot u_x(\lambda x, \lambda y) \right] = u_x(\lambda x, \lambda y)$$ $$v_{xx} = \frac{\partial}{\partial x} \left[ u_x(\lambda x, \lambda y) \right] = \lambda u_x(\lambda x, \lambda y)$$ $$v_y = \frac{\partial}{\partial y} \left[ \lambda^{-1} u(\lambda x, \lambda y) \right] = \frac{1}{\lambda} \left[ \lambda \cdot u_y(\lambda x, \lambda y) \right] = u_y(\lambda x, \lambda y)$$ $$v_{yy} = \frac{\partial}{\partial y} \left[ u_y(\lambda x, \lambda y) \right] = \lambda u_y(\lambda x, \lambda y)$$ $$v_{xy} = \frac{\partial}{\partial y} \left[ u_x \right] = \lambda u_{xy}(\lambda x, \lambda y)$$
Then, by plugging these into the original Minimal Surface Equation, we can obtain: $$(1 + v_y^2)v_{xx} + (1+v_x^2)v_{yy} - 2v_xv_yv_{xy}=0$$ $$(1 + [u_y(\lambda x, \lambda y)]^2)[\lambda u_x(\lambda x, \lambda y)] + (1 + [u_x(\lambda x, \lambda y)]^2)[\lambda u_y(\lambda x, \lambda y)] \\ - 2 [u_x(\lambda x, \lambda y)] [ u_y(\lambda x, \lambda y)] [\lambda u_{xy}(\lambda x, \lambda y)]$$
Factoring out $\lambda$ gives the original equation, but evaluating $u$ at $(\lambda x, \lambda y)$ for every point. While conceptually this seems right, how can I verify that this will still satisfy the original PDE?
Note: This is part of an assignment, and as such, I do not want a blatant answer to my problem. I checked the syllabus and it encourages working on the homework with peers, so I believe this is allowed, especially since I've already done the majority of the work for the problem. If someone could point me in the right direction, or help me find the trick that I'm missing, I'd appreciate it.