Answer
Let $y=f(x)$ be a function whose graph is symmetric about y-axis as well as origin.
Now, the graph which is symmetric about y-axis has a property that if $(x, y)$ is a point on the graph, then, $(-x, y)$ is also a point on the graph. Thus, for our function to be symmetric about y-axis, we must have $f(-x)=f(x)$[Equation 1] for all $x$.
Also, the graph symmetric about origin follows that if $(x,y)$ is a point on the graph, then, we must have $(-x, -y)$ on the graph as well. Thus, for our function to be symmetric about origin, we must have $f(-x)=-f(x)$[Equation 2] for all $x$.
Now, from Equation 1 and Equation 2, we have
$f(x)=-f(x) \Rightarrow 2f(x)=0$, or, $f(x)=0$ for all $x$.
Therefore, the only function which is symmetric with respect to both y-axis and origin is $f(x)=0$ .
Work Step by Step
Note that $f(x)$ denotes the height above $x$, or, the y-coordinate for a graph of the function, thus, $f(-x)$ means height above $-x$.
