Local max/min of this $$f(x,y)=(x-y)^n$$ where $n \geq 1$ 
using this limitation $ x^2 + y^2 -1 =0$
I tried to find the derivatives 
$$
\begin{cases}
f_x = 2λx + n(x-y)^{n-1}=0 \\
f_y = 2λy - n(x-y)^{n-1}=0\\
f_λ = x^2 + y^2 -1 =0
 \end{cases}
$$
I get $x=y \quad λ=0 \quad with \quad x=\pm\sqrt{\frac{1}{2}}$
However I have been told that this is wrong.And then 3x3 matrix becomes complex.But suggested (0,0,?) solution doesnt make sense as $f_l$ cant be zero 

$(1)$ Hint: $$ (x+(-y))^n = \sum_{k=0}^{n} {n\choose k} x^{n-k}(-y)^k$$
$(2)$ $$x^2+y^2 = 1 \Rightarrow x = \sqrt{1-y^2} \Rightarrow (x-y)^n = (\sqrt{1-y^2}-y)^n.$$ $$\text{Now find the zeros of} \ \ g(x)=\sqrt{1-y^2}-y.$$ 
$$\text{Verify that your zero is the point $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$}. $$
$\textbf{Comment}$: And this makes sense because any zero of $f(x,y)=(x-y)^n$ is when $x=y$ and if we are limited to points on the boundary of the unit circle, the only time this happens is at the angle $\frac{\pi}{4}$. Now we understand this a bit better and it should follow that the max of $f(x,y)$ with this limitation is $1$ and the min is $-1$.
