Does there exist a continuous function, which is symmetric about $y=-x$, $x\in[-\infty,\infty]$?
In addition, I wish the graph is a curve but not a straight line, i.e., $f′(x)$ is not a constant.
In my opinion, there is no such curve. Because if it exists, there will be more than two y(x) for each x except the points on $y=-x$.
If you don't want just a straight line, one approach, suggested by Rahul in comments, is to stitch the function together from two halves. The northeast half could be something like $x^2$ which we know how to invert, but we need to lower it by $\frac34$ such that it will meet the $x+y=0$ line at a right angle so the derivative will be continuous. To the southwest we simply mirror the parabola we already have: $$ f(x) = \begin{cases} x^2-3/4 & \text{when }x\ge1/2 \\ -\sqrt{3/4-x} & \text{when }x\le1/2 \end{cases} $$
This is not quite smooth at the point where it crosses the antidiagonal, though. You can make a smoother result by taking an even analytic function without too large derivatives, such as $$ g(x) = \frac12\cos x $$ and geometrically rotating its graph it by $45^\circ$ about the origin. The resulting curve will satisfy the conditions for being a the graph of a function, and be as smooth as you want. But it will usually not have a nice algebraic expression.
One case where we do get a single algebraic expression is if instead we take $g$ to be $g(x)=\sqrt{1+x^2/2}$. Its graph is the upper branch of the hyperbola $$ y^2 = 1 + \tfrac12 x^2 $$ and turning this by $45^\circ$ gives the equation $$ \biggl(\frac{y-x}{\sqrt2}\biggr)^2 = 1 + \frac12\biggl(\frac{x+y}{\sqrt2}\biggr)^2 $$ which we can rearrange into a quadratic equation in $y$ and solve to get $$ y = 3x \pm 2\sqrt{1+2x^2} $$ So $f(x) = 3x + 2\sqrt{1+2x^2}$ is a completely smooth function whose graph is symmetric about the line $x+y=0$ (even though it is not obvious from the formula that this is the case).