$x^2+\frac{x^2}{(x+1{)}^2}=3$

The given equation is

$\begin{array}{l}{x}^2+\frac{x^2}{(x+1{)}^2}=3\\ x^4+2x^3-x^2-6x-3=0\end{array}$

Let fix)= x⁴ + 2x³ -x² - 6x - 3

f(x):++--- 

f(-x): +--+- 

In f(x), there are one change, so f(x) has one +ve root and in f(-x), there are 3 changes, so f(-x) has 3 negative roots Thus, the number of real roots = 1 + 3 = 4