= Solution
The assertion is \b[false]. Use the <pumping lemma for regular languages>: if $L$ is a <regular language>, there is $p\ge1$ such that every $w\in L$ with $|w|\ge p$ admits a factorization $w=xyz$ with $|xy|\le p$, $|y|\ge1$, and $xy^kz\in L$ for every integer $k\ge0$.
Suppose the stated <formal language> were a <regular language>, and take $w=0^{2p}1^p$. This <string> has twice as many zeros as ones, so it belongs to $L$. In every allowed factorization, the first $p$ positions are all zeros, forcing $y=0^d$ with $1\le d\le p$. Pumping down with $k=0$ produces
$$
xz=0^{2p-d}1^p.
$$
It still has $p$ ones but fewer than $2p$ zeros, contradicting the <pumping lemma for regular languages>. Thus
$$
\boxed{\{w:|w|_0=2|w|_1\}\text{ is not regular}.}
$$
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