Step-by-step Solution:

Let \[ B=\sin^{2}y+\cos^{4}y. \] Put \(s=\sin^{2}y\). Then \(0\le s\le 1\) and \(\cos^{2}y=1-s\). Hence \[ B=s+(1-s)^{2}=s+1-2s+s^{2}=s^{2}-s+1. \] Consider the quadratic \(f(s)=s^{2}-s+1\) on the interval \([0,1]\). Its vertex is at \(s=\tfrac{1}{2}\), and \[ f\!\left(\tfrac{1}{2}\right)=\tfrac{1}{4}-\tfrac{1}{2}+1=\tfrac{3}{4}. \] Also \(f(0)=1\) and \(f(1)=1\). Therefore for all real \(y\), \[ \boxed{\tfrac{3}{4}\le B\le 1.} \] Equality cases: \(B=\tfrac{3}{4}\) when \(\sin^{2}y=\tfrac{1}{2}\) (e.g. \(y=\tfrac{\pi}{4}+k\frac{\pi}{2}\)), and \(B=1\) when \(\sin^{2}y=0\) or \(1\) (i.e. \(y=k\pi\) or \(y=\tfrac{\pi}{2}+k\pi\), \(k\in\mathbb{Z}\)).