We have, $f(x)$ is a quadratic equation.
$$
\begin{array}{rlrl}
& \therefore \quad f(x)=a x^2+b x+c \\
& f(0) & =c, f(1)=a+b+c \\
& & f(0)+f(1) & =0 \\
\Rightarrow & c+a+b+c & =0 \\
\Rightarrow \quad & a+b+2 c & =0
\end{array}
$$


$$
\begin{aligned}
f(-2) & =a(-2)^2+b(-2)+c \\
0 & =4 a-2 b+c \\
\Rightarrow \quad 4 a-2 b+c & =0
\end{aligned}
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
& \frac{a}{5}=\frac{b}{7}=\frac{c}{-6} \\
& \Rightarrow \quad a=5 k, b=7 k, c=-6 k \\
& \therefore \quad f(x)=k\left(5 x^2+7 x-6\right) \\
& \Rightarrow 5 x^2+7 x-6=0 \\
& \Rightarrow \quad 5 x^2+10 x-3 x-6=0 \\
& \Rightarrow \quad 5 x(x+2)-3(x+2)=0 \\
& \Rightarrow \quad(x+2)(5 x-3)=0 \\
& x=-2, x=\frac{3}{5} \\
&
\end{aligned}
$$
Hence, $f\left(\frac{3}{5}\right)=0$.