minimum of this function Define $f(x,y)  =  (a - bx - by)e^{-(x^2+c y^2)}$, where $c > 1$ and $b>a$ then define
$g(x,y)  =  f(x,y)$ if $x<y$ and $g(x,y)=f(y,x)$ if $x>=y$.
Look at the function $g(x,y)$.
Obviously $g(x,y)$ is symmetric about $x=y$ and not differentiable at $x=y$, but I don't know why the minimum of $g(x,y)$ also occurs at $x=y$?
 A: Assume: $$a,b,c,x,y\in\mathbb{R}:\,c>1,$$
and define:
$$g:\mathbb{R}^2\rightarrow\mathbb{R}.$$
Change of coordinates
Consider: $$x=\frac{1}{2}(v+u),\,\,y=\frac{1}{2}(u-v),$$ $$u=x+y,\,\,v=x-y,$$ $$\displaystyle g(u,v)=\left( a-bu \right) \,{\mathbb{exp}\left[{-\frac{1}{4}\, \left( {v}^{2}+{u}^{2} \right) 
 \left( 1+c \right) +\frac{1}{2}\, \left( 1-c \right)\, u \left| v \right| }\right]}.
$$
The absolute value of $v$ takes care of the piecewise definition, highlights the symmetry and shows that the function is not differentiable w.r.t $v$ at $v=0$. 
Are there stationary values along $v\ne0$?
$${\frac {\partial }{\partial v}}g \left( u,v\neq 0 \right) =0\Rightarrow \left| v \right| ={\frac { \left( 1-c \right) u}{1+c}}$$
For $c>1\, (c<1)$ there are stationary points along $v$, symmetrically distributed about $v=0$, when $u<0\, (u>0)$; that is the only region in which the above equation is solvable. These stationary points coalesce as $u\rightarrow 0$. Where the above equation is not solvable there can be no stationary points at $v\ne0$ and thus the maximum or minimum of $g$ must lie at the discontinuity $v=0$.
Are the stationary points at $v\ne0$ maxima or minima?
For $c>0$ the exponential decays monotonically to zero for large $v$, this, the quadratic nature of the $v$ dependence and the positivity of the exponential term, insure that, where the stationary points exist:


*

*the value of $g$ at the discontinuity $v=0$, is closer to $0$ than the value of $g$  at the stationary points,

*the stationary points are maxima along $v$ when $a-bu>0$ and are minima when $a-bu<0$.


When are the conditions for stationary points existing at $v\ne0$ and a negative pre-factor simultaneously solvable?
Define the prefactor: $$p=a-ub=|b|\left(\mathbb{sign}(a)\left|\frac{a}{b}\right|-\mathbb{sign}(b)u\right)$$
and examine the table below.
\begin{array}{|c|c|c|c|} \hline \mathbb{sign}(a)& \mathbb{sign}(b) & \text{cond. on } u\,\, \text{for}\,\, p<0 & \text{min/max} \\ \hline+&+&u>\left|\frac{a}{b}\right|&\text{1 min @}\,v=0,\text{2 max @}\,v\ne0\, \\ \hline +&-&u<-\left|\frac{a}{b}\right|&\text{2 min @}\,v\ne0,\text{1 max @}\,v=0\, \\ \hline -&+&u>-\left|\frac{a}{b}\right|&\text{1 min @}\,v=0,\text{2 max @}\,v\ne0\,  \\ \hline -&-&u<\left|\frac{a}{b}\right|&\text{2 min @}\,v\ne0,\text{1 max @}\,v=0\,\\ \hline \end{array}
For $p<0$ the first two rows force $u$ to be exclusively $+$ or $-$ and the final column is filled in accordingly from previous discussions. In the second two rows $u$ is not required to be exclusively $+$ or $-$ but these are filled in by recognising them as simple over all sign changes in $g$ for rows $1$ and $2$. In the final row, we see that $a<b$ is not sufficient for one minimum at $v=0$ if $b<0$. 

Rather, a necessary and sufficient condition
for a single minimum at $v=0$ is: $$0<b,$$
and you then have two maxima at $v\ne0$.

Where are the stationary points located?
The min/max stationary point @ $v=0$ is found using:
$${\frac {\partial }{\partial u}}g \left( u,0 \right) ={u}^{2}-{\frac {a
u}{b}}-\, \frac{2}{1+c}
=0,$$ $$u=\frac{1}{2}\,{\frac {a}{b}}\pm\frac{1}{2}\,\sqrt {{\frac {{a}^{2}}{{b}^{2}}}+
 \frac{8}{1+c}},$$
and the conditions in the table dictate that, as $0<c$ (in fact $c>1$), we must chose the $+$ in all cases and thus, in all cases this stationary point is located at:
$$u=\frac{1}{2}\,{\frac {a}{b}}+\frac{1}{2}\,\sqrt {{\frac {{a}^{2}}{{b}^{2}}}+
 \frac{8}{1+c}},\,v=0\,:\,x=y=\frac{1}{4}\,{\frac {a}{b}}+\frac{1}{4}\,\sqrt {{\frac {{a}^{2}}{{b}^{2}}}+
 \frac{8}{1+c}}.$$
In the above, the $-$ choice corresponds to a saddle-like point between the two min/max stationary points @ $v\ne0$. The min/max stationary points @ $v\ne0$ themselves can be located using:
$$\left({\frac {\partial }{\partial u}}g \left( u,v \right)\right)|_{ 
 \left| v \right| ={\frac { ( 1-c ) u}{1+c}}} =0\Rightarrow {u}^{2}-{\frac {a}{b}}u-\frac{1}{2c}-\frac{1}{2}$$
$$u={\frac {a}{2b}}\pm\frac {1}{2}\,\sqrt {{\frac {{a}^{2}}{{b}^{2}}}+\frac {2}{c}
+2}$$
and because the existence of these points requires $u<0$ we take the $-$ and we then have the location of two stationary points at:
$$u={\frac {a}{2b}}-\frac {1}{2}\,\sqrt {{\frac {{a}^{2}}{{b}^{2}}}+\frac {2}{c}
+2},\,\,v=\pm\frac{1-c}{1+c}\left({\frac {a}{2b}}-\frac {1}{2}\,\sqrt {{\frac {{a}^{2}}{{b}^{2}}}+\frac {2}{c}
+2}\right)$$
Example:
$$a=1,b=2,c=3,$$
minimum of $g$ at: $$u=1,v=0:x=y=\frac{1}{2}.$$

A: (not a complete answer, but a start.)
The partial of $f(x,x+t)$ with respect to $t$ at $t=0$ is
$$(4bcx^2-2acx-b)e^{(-c-1)x^2}.$$ 
The quadratic factor here has positive discriminant, so to know for sure one has a max or a min along $y=x$ some assumption about $a,b,c$ must force a certain sign for the quadratic factor. At present I don't see how your conditions $c>1,b>a$ alone decide the sign on the quadratic.
