For instance, choose any point $p$ on the line $x-y=1$, which we can denote as $(c+1,c)\in\mathbb{R}^2$.
Regard the output of $f$ as $(f_1,f_2)$, where $f_1(x,y)=x^3+3xy^2-15x-12y$ and $f_2(x,y)=x+y$.
Our arbitrarily small neighbourhood of $p$, we can call $U$.
We are looking for points $p_1=(x_1,y_1)$ and $p_2=(x_2,y_2)$, distinct points in $U$, for which the output of $f$ is the same.
Let's try having at least $f_2(x_1,y_1)=f_2(x_2,y_2)$, since $f_2$ is easier than $f_1$. This is equivalent to $x_1+y_1=x_2+y_2$. Both points therefore belong to the same line $L:x+y=k$. To make things even simpler, I'll adjust $k$ so that $p$ is on that line, i.e. $k=(c+1)+c=2c+1$.
This allows us to substitute $y=2c+1-x$ for any arbitrary point on $L$. This in turn allows us to parametrise the points on $L$. The $x$-coordinate uniquely identifies each point on $L$, and the corresponding $y$-coordinate is $2c+1-x$.
Plug that straight into $f_1$. One will get $f_1(x,2c+1-x)=\cdots$ I can't bother writing it out.
But using Wolfram Alpha, you can check that $\dfrac{d}{dx}f_1(x,2c+1-x)=12 (c + c^2 - 2 c x + (-1 + x) x)$, which is equal to $0$ when $x$ is equal to $c+1$. Furthermore, $\dfrac{d^2}{dx^2}f_1(x,2c+1-x)=-12(2c-2x+1)$, which is equal to $12$ when $x$ is equal to $c+1$.
Therefore, $f_1(x,2c+1-x)$ actually attains a local maximum at $x=c+1$.
No matter how small $U$ is,
you can find $x_1$ slightly smaller than $c+1$, and $x_2$ slightly larger than $c+1$,
such that $f_1(x_1,2c+1-x_1)=f_1(x_2,2c+1-x_2)$.
Then we have our choice of $(x_1,y_1)$, where $y_1=2c+1-x_1$. Likewise for $x_2$.
