Let us use the following lemma :
Lemma : To find the maximum value of $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor$, we only need to consider $(x,y)$ such that $(x-2)^2+(y-2)^2=250$.
The proof of the lemma is written at the end of the answer.
From the lemma, we have
$$y=2-\sqrt{250-(x-2)^2}$$
which is decreasing for $x\lt 0$.
We may suppose that $x\ge y$. Now, solving the system
$$250-(x-2)^2\ge 0\qquad\text{and}\qquad x\lt 0\qquad\text{and}\qquad y\lt 0\qquad\text{and}\qquad x\ge y$$
gives
$$(-9.18\approx)\ 2 - 5 \sqrt 5\le x\lt 0$$
If $-1\le x\lt 0$, then we get$$y\gt 2-\sqrt{250-(0-2)^2}=2-\sqrt{246}\gt 2-\sqrt{246.49}=2-15.7=-13.7$$Since $\lfloor x\rfloor=-1$, we get $x\lfloor x\rfloor\le 1$ implying $\lfloor x\lfloor x\rfloor\rfloor\le 1$. Since $\lfloor y\rfloor\ge -14$, we get $y\lfloor y\rfloor\lt (-14)\times (-13.7)=191.8$ implying $\lfloor y\lfloor y\rfloor\rfloor\le 191$. So, $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 1+191=192$
If $-2\lt x\lt -1$, then $-13.53\lt y$, so $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 3+189=192$
If $-\frac 73\lt x\le -2$, then $-13.3\lt y$, so $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 6+186=192$
If $-3\lt x\le -\frac 73$, then $-13.21\lt y$, so $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 8+184=192$
If $-4\lt x\le -3$, then $-13\le y$, so $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 15+169=184$
If $-5\le x\le -4$, then $-12.7\lt y$, so $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 25+165= 190$
If $-\frac{17}{3}\le x\lt -5$, then $-12.2\lt y$, so $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 34+158= 192$
If $-7\lt x\lt -\frac{17}{3}$, then $-11.9\lt y$, so $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 48+142=190$
If $-8\le x\le -7$, then $-11\le y$, so $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 64+121=185$
If $-\frac{79}{9}\le x\lt -8$, then $-10.3\lt y$, so $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 79+113=192$
If $-9\le x\lt -\frac{79}{9}$, then $-9.6\lt y$, so $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 81+96=177$
If $2-5\sqrt 5\le x\lt -9$, then $-9.4\lt y$, so $\lfloor x\lfloor
x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 91+94=185$
It follows from these that
$$\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor\le 192$$
whose equality is attained when
$$(x,y)=\bigg(-\frac{201}{100},\frac{200-9\sqrt{28879}}{100}\bigg)$$
Therefore, the maximum value of $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor$ is $\color{red}{192}$.
Finally, let us prove the lemma.
Lemma : To find the maximum value of $\lfloor x\lfloor x\rfloor\rfloor+\lfloor y\lfloor y\rfloor\rfloor$, we only need to consider $(x,y)$ such that $(x-2)^2+(y-2)^2=250$.
Proof for lemma :
Let $m$ be a negative integer. Also, let $\alpha$ be a real number such that $0\le \alpha\lt 1$.
To prove the lemma, it is sufficient to prove the followings :
(1) For any fixed $m$, $f(\alpha):=\lfloor (m+\alpha)\lfloor (m+\alpha)\rfloor\rfloor$ is decreasing.
(2) For any $(m,\alpha)$, $\lfloor (m+\alpha)\lfloor m+\alpha\rfloor\rfloor\ge \lfloor (m+1)\lfloor m+1\rfloor\rfloor$
Proof for (1) : $$f(\alpha)=\lfloor (m+\alpha)\lfloor (m+\alpha)\rfloor\rfloor=m^2+\lfloor m\alpha\rfloor$$is decreasing.
Proof for (2) :
$$\begin{align}&\lfloor (m+\alpha)\lfloor m+\alpha\rfloor\rfloor- \lfloor (m+1)\lfloor m+1\rfloor\rfloor
\\\\&=m^2+\lfloor m\alpha\rfloor-(m+1)^2
\\\\&=-2m-1+\lfloor m\alpha\rfloor
\\\\&\ge -2m-1+m
\\\\&\ge 0\qquad\square\end{align}$$
Best Answer
You have correctly deduced that $\color{blue}{0\leq x<\frac 53}$. To get the solution, the given equation $4\lfloor x\rfloor=x+\{x\}$ must also be satisfied.
Note that $x=1$ does not satisfy the given equation. If $x>1$, then $\lfloor x\rfloor=1$ and hence $4=x+\{x\}<\frac 53+1=8/3$ (by the blue colored inequality) , which is not possible.
Hence it follows that $0\leq x<1$.
Since, $\lfloor x\rfloor=0$, it follows that $\color{red}{x+\{x\}=0}$ (due to the given equation).
The identity $x=\lfloor x\rfloor+\{x\}$ gives $x=\{x\}$, which along-with the red-colored equation gives: $x=0$.