Suppose $Df(z)^TDf(z) = \lambda(z)I$. Show $f(z)$ is holomorphic or $\overline{f(z)}$ is holomorphic From an old qualifier: 

Let $z=x+iy$, $f=f(z)=u+iv$. Assume $\Omega$ is an open connected
  domain in $\mathbb{C}$, $f\in C^2(\Omega)$. Denote $$Df = \left[ 
 \begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial
 u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial
 v}{\partial y} \end{matrix} \right].$$ Suppose for every $z\in
 \Omega$, $$Df(z)^TDf(z) = \lambda(z)I$$ for some $\lambda(z)$, where
  $I$ is the 2x2 identity matrix. Then show that either $f(z)$ is
  holomorphic or $\overline{f(z)}$ is holomorphic.

Ideas: We get $$\left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial v}{\partial x} \right)^2 = \left( \frac{\partial u}{\partial y} \right)^2 + \left( \frac{\partial v}{\partial y} \right)^2 \quad \text{and}\tag{1}$$ $$\frac{\partial u}{\partial x}\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\frac{\partial v}{\partial y} =0\tag{2}.$$
What I've been doing is differentiating (1) and (2) with respect to $x$ and $y$ and trying to cancel. One thing I ended up with is $$\Delta u \left(\frac{\partial u}{\partial x} - \frac{\partial u}{\partial y}\right) + \Delta v \left(\frac{\partial v}{\partial x} - \frac{\partial v}{\partial y}\right)=0.$$
But I need to show relations between $\partial u/\partial x$ and $\partial v/\partial y$ etc.
 A: If $A^TA = \lambda I$, then also $AA^T = \lambda I$. Using that, we have
$$Df(z)Df(z)^T = \lambda(z) I,$$
and that translates to
$$\begin{gather}
u_x^2 + u_y^2 = v_x^2 + v_y^2,\\
u_x v_x + u_y v_y = 0.
\end{gather}$$
So the gradients of $u$ and $v$ have equal length everywhere, and are orthogonal, hence
$$\begin{pmatrix} v_x\\v_y\end{pmatrix} = \pm \begin{pmatrix}0&-1\\1&0 \end{pmatrix}\cdot \begin{pmatrix}u_x\\u_y \end{pmatrix},$$
and that means in each point either $f$ or $\overline{f}$ satisfies the Cauchy-Riemann equations.
A: What might be termed the "transpose" of Daniel Fischer's argument:
Writing out $DF(z)^TDf(z)$ with
$Df = \left[ 
 \begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial
 u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial
 v}{\partial y} \end{matrix} \right], \tag{1}$
as it is defined by the OP Eric Auld, we arrive at a slight modification of his equations (1), (2), viz.
$\left( \frac{\partial u}{\partial x} \right)^2 + \left( \frac{\partial v}{\partial x} \right)^2 = \left( \frac{\partial u}{\partial y} \right)^2 + \left( \frac{\partial v}{\partial y} \right)^2 = \lambda(z) \tag{2}$
and
$\frac{\partial u}{\partial x}\frac{\partial u}{\partial y} + \frac{\partial v}{\partial x}\frac{\partial v}{\partial y} =0\tag{2}.$
Setting $u_x = \partial u / \partial x$, etc., i.e. using the subscript notation for partial derivatives, we then have
$u_x^2 + v_x^2 = u_y^2 + v_y^2 = \lambda(z) \tag{4}$
and 
$u_x u_y + v_x v_y = 0. \tag{5}$
Consider the vector fields $(u_x, v_x)$ and $(u_y, v_y)$.  By (4) they are of the same magnitude, and by (5), they are orthogonal.  Thus $(u_x, v_x)$ must be a scalar multiple of
$(v_y,- u_y)$.  By (4), we must then have $(u_x, v_x) = \pm(v_y,- u_y)$; the "$+$" sign gives the Cauchy-Riemann equations for $f(z)$; the "$-$" sign for $\overline{f(z)}$.  Thus either $f(z)$ or $\overline{f(z)}$ is holomorphic.
Hope this helps.  Happy New Year,
and as always,
Fiat Lux!!!
A: From (2), $\dfrac{u_x}{v_y}= -\dfrac{v_x}{u_y} = k$(say). (Justification for $u_y\neq0$ and $v_y\neq 0$ later).
Using in (1) ; $(k^2-1)(v_y^2 +u_y ^2) = 0  \implies  k = \pm 1$. Hence either $f$ or $\bar{f}$ is holomorphic. 
If $u_y = v_y =0$, then
$Df^TDf = \left[ 
 \begin{matrix} u_x^2 + v_x^2 & 0 \\ 0 & 0 \end{matrix} \right] = \lambda(z) I$ does not make sense. So both cannot be zero. 
If one is zero (WLOG) $u_y =0$: then $v_xv_y = 0$. This means $v_x = 0 \implies u_y = v_x$. Also, $Df^TDf = \left[ 
 \begin{matrix} u_x^2  & 0 \\ 0 & v_y^2 \end{matrix} \right] = \lambda(z) I \implies u_x = \pm v_y$.
A: Consider the surface $M \subseteq \mathbb R^2$ parameterized by $$f(x,y)=(u(x,y),v(x,y)). $$ Your condition implies that the first fundamental form associated with this map is scalar, which is also called conformal. Conformal maps preserve angles, and all that is needed to finish is the relationship between conformality and holomorphicity, which is discussed in introductory complex analysis texts, e.g. Ahlfors [p. 73].
