Help to clarify proof of Euler's Theorem on homogenous equations 
Why is the last step (setting $\lambda = 1$ allowed?
I have trouble accepting this because if I set $\lambda =1 $ at the very start, then:
$f(\lambda x , \lambda y)=\lambda^r f(x,y)$
becomes
$f(x,y) = f(x,y)$
and
so I can't prove the theorem.
Why does setting $\lambda = 1$ at the end of the proof work?
EDIT
For example, let $f(x,y)=x^2y^2$
Therefore, $f(x,y)$ is homogeneous of degree 4.
Therefore, the second to last step will be:
$4x^2y^2=4\lambda^3x^2y^2$
The above equation is only true if $\lambda=1$
Why doesn't the theorem make a qualification that $\lambda$ must be equal to 1?
It seems to me that this theorem is saying that there is a special relationship between the derivatives of a homogenous function and its degree but this relationship holds only when $\lambda=1$. Please correct me if my observation is wrong.
 A: You can't set $\lambda = 1$ in the line $f(\lambda x,\lambda y) = \lambda^r f(x,y)$ because the very next step is to differentiate with respect to $\lambda$ which makes no sense when the variable isn't present.
At the end of the proof, you are taking advantage of the fact that $\lambda$ is an arbitrary element of $\mathbb{R}$. So you pick an element that makes the calculations easy.
Variant of proof
Define the function $g : \mathbb{R} \rightarrow \mathbb{R}$ by $g(t) = f(tx,ty)$. Since $f$ is homogeneous, we can write $g(t) = t^r f(x,y)$. Find $g'(t)$. 
Using $g(t) = t^r f(x,y)$, it is clear that $g'(t) = rt^{r-1} f(x,y)$.
Using $g(t) = f(tx,ty)$, we get that $g'(t) = \frac{\partial f}{\partial (tx)}\cdot\frac{d(tx)}{dt} + \frac{\partial f}{\partial (ty)}\cdot\frac{d(ty)}{dt} = x\frac{\partial f}{\partial (tx)}+y\frac{\partial f}{\partial (ty)}$.
So we have that for all $t$, $rt^{r-1} f(x,y) = x\frac{\partial f}{\partial (tx)} + y\frac{\partial f}{\partial (ty)}$. If we let $t=1$, then we have that $g(1) = f(x,y)$, our original function, and $rf(x,y) = x\frac{\partial f}{\partial x}+ y\frac{\partial f}{\partial y}$, the desired result.

To address the specific example of $f(x,y) = x^2y^2$. You have the second to last step wrong on the RHS. 
When doing the proof, we are working with $f(\lambda x, \lambda y)$ throughout. So when calculating $\dfrac{df}{d\lambda}$ we get:
$\begin{align}
\dfrac{df}{d\lambda} &= \dfrac{\partial f}{\partial (\lambda x)}\cdot \dfrac{d (\lambda x)}{d\lambda} + \dfrac{\partial f}{\partial (\lambda y)}\cdot \dfrac{d (\lambda y)}{d\lambda} \\
&= 2(\lambda x)(\lambda y)^2\cdot x + 2(\lambda x)^2(\lambda y)\cdot y \\
&= 4\lambda^3 x^2y^2 = \dfrac{d}{d\lambda}\left(\lambda^4 f(x,y)\right)
\end{align}$
So we could choose any $\lambda$ we want and it would still be a true equation. But then to get the desired result, we would have to divide that back out of both sides. Choosing $\lambda = 1$ saves a bit of algebra.
A: Regarding your example: In the second to last step where you compute
$$
\frac{\partial f}{\partial u} \frac{\mathrm du}{\mathrm d\lambda}
+ \frac{\partial f}{\partial v} \frac{\mathrm dv}{\mathrm d\lambda}
= r\lambda^{r-1}f(x,y)
$$
In your example, the left-hand side of that is computed as follows: We differentiate $f(\lambda x, \lambda y)$ with respect to $\lambda$. To do that, we use the chain rule: The outer derivative is the derivative of $f$ evaluated at $(\lambda x, \lambda y)$, the inner derivative is just $(x,y)$. So we need to compute (I'm writing this as a product of vectors here)
$$ Df(\lambda x, \lambda y) \cdot (x,y) $$
Since $Df(x,y) =  (2xy^2, 2x^2y)$, this gives
$$ Df(\lambda x, \lambda y) \cdot (x,y) = (2(\lambda x)(\lambda y)^2, 2(\lambda x)^2(\lambda y)) \cdot (x,y) = (2\lambda^3 xy^2, 2\lambda^3 x^2y) \cdot (x,y) = 2 \lambda^3 x^2y^2 + 2 \lambda^3 x^2y^2 = 4 \lambda^3 x^2y^2$$
as desired.
A: I can say that the general property which must be satisfied for all $\lambda>0$ is :
$$x\frac{\partial f}{\partial x}(\lambda x,\lambda y)+y\frac{\partial f}{\partial y}(\lambda x,\lambda y)=k\lambda^{k-1}f(x,y).$$
From which we deduce the desired Euler Theorem by setting $\lambda=1$.
A: You can derive Euler theorem without imposing $\lambda=1$.
Starting from $f(\lambda x, \lambda y) = \lambda^n \times f(x,y)$, one can write the differentials of the LHS and RHS of this equation:

*

*LHS

$df(\lambda x, \lambda y) = \left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} \times d(\lambda x) + \left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} \times d(\lambda y)$
One can then expand and collect the $d(\lambda x)$ as $xd\lambda + \lambda dx$ and $d(\lambda y)$ as $yd\lambda + \lambda dy$ and achieve the following relation:
$df(\lambda x, \lambda y) = \lambda \left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} \times dx + \lambda \left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} \times dy + \left[x\left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} + y\left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x}\right] \times d\lambda$

*

*RHS

$d\left[\lambda^n f(x,y)\right] = n\lambda^{n-1}f(x,y) \times d\lambda + \lambda^n \left[\left(\frac{\partial f}{\partial x}\right)_y \times dx + \left(\frac{\partial f}{\partial y}\right)_x \times dy\right]$

*

*RHS = LHS

Both differentials must be equal. Thus, terms in front of $dx$, $dy$ and $d\lambda$ must be equal in both expressions. This leads to the following system:
$\begin{cases}
\lambda \left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} = \lambda^n \times \left(\frac{\partial f}{\partial x}\right)_y\\
\lambda \left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} = \lambda^n \times \left(\frac{\partial f}{\partial y}\right)_x\\
x\left(\frac{\partial f}{\partial \lambda x}\right)_{\lambda y} + y\left(\frac{\partial f}{\partial \lambda y}\right)_{\lambda x} = n \lambda^{n-1} \times f(x,y)
\end{cases}$
From this point, the classical next step is to take the third equation, set $\lambda=1$ (although you want the identity to be true for all $\lambda$) and conclude that the identity is proven. Clearly this assumption is not required as you can inject the first and second equations into the third and obtain the required identity:
$\lambda^{n-1} x\left(\frac{\partial f}{\partial \lambda x}\right)_{y} + \lambda^{n-1} y\left(\frac{\partial f}{\partial \lambda y}\right)_{x} = n \lambda^{n-1} \times f(x,y)$
You can see that all $\lambda^{n-1}$ cancel and lead to the expected relation:
$x\left(\frac{\partial f}{\partial \lambda x}\right)_{y} + y\left(\frac{\partial f}{\partial \lambda y}\right)_{x} = n \times f(x,y)$
A: Consider a function $f$ that is homogeneous of degree $r$, i.e.,
$f(tx, ty) = t^rf(x, y)$ ......$(1)$
Differentiate both sides of equation (1) with respect to $t$. Then by Chain Rule:
$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)}x + \frac{\partial f(tx, ty)}{\partial(ty)}y = rt^{r-1}f(x, y) 
\end{equation}$ ......$(2)$
Differentiate both sides of equation $(1)$ with respect to $x$. Then by chain rule:
$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)}t = t^r\frac{\partial f(x, y)}{\partial(x)} 
\end{equation}$
$\implies$$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(tx)} = t^{r-1}\frac{\partial f(x, y)}{\partial x} 
\end{equation}$ ......$(3)$
Similarly,
$\begin{equation}
\frac{\partial f(tx, ty)}{\partial(ty)} = t^{r-1}\frac{\partial f(x, y)}{\partial y} 
\end{equation}$ ......$(4)$
Put $(3)$ and $(4)$ in $(2)$ to get the result:
$\begin{equation}
\frac{\partial f(x, y)}{\partial x}x + \frac{\partial f(x, y)}{\partial y}y = rf(x, y) 
\end{equation}$
A: Since $f(x,y)=x^2y^2$, therefore, it can be written as
$$f(x,y)=x^2\left(\frac{y}{x}\right)\times x^2=x^4\left(\frac{y}{x}\right).$$
This shows that $f$ is a homogeneous function of degree $4$. Hence, by Euler's theorem, we have
$$x\frac{\partial f}{\partial x} + x\frac{\partial f}{\partial x}=4f.$$
