smc

Let $y=x^2+1$. Then $x^2=y-1$, so we can write the given equation as $$\begin{array}{rl}f(y)& =x^4+5x^2+3\\ & =(x^2{)}^2+5x^2+3\\ & =(y-1{)}^2+5(y-1)+3\\ & =y^2-2y+1+5y-5+3\\ & =y^2+3y-1.\end{array}$$ Then substituting $x^2-1$ for $y$, we get $$\begin{array}{rl}f(x^2-1)& =(x^2-1{)}^2+3(x^2-1)-1\\ & =x^4-2x^2+1+3x^2-3-1\\ & =x^4+x^2-3.\end{array}$$ The answer is therefore $\boxed{x^4+x^2-3}$.