The answer is no. Suppose otherwise and let $PQR$ be the regular triangle inscribed in the regular hexagon $A_1A_2A_3A_4A_5A_6$. Extend lines $PQ, QR, RP$ infinitely and observe that the region from where all three vertices can be seen is precisley the union of the $60^{\circ}$ regions subtended at the vertices $P, Q, R$ respectively. In particular, if we were to squish the triangle into it's barycenter while preserving the slope of the regions we would have a point $O$ strictly inside the regular hexagon such that each vertex of the hexagon lies in "alternating" regions among the $6$ regions created by $O$. Observe that there are three such regions, so one of them contains at least two. If one of them contains $3$, then we'll find two consecutive vertices of the hexagon subtending an angle of $ < 60^{\circ}/2 = 30^{\circ}$ at $O$. But this would push $O$ outside the circumcircle of the hexagon, a contradiction. Thus each reigon has exactly $2$ points. Suppose $(A_1, A_2);(A_3, A_4);(A_5, A_6)$ are the pairs of points in the same region. Now observe that we have $\angle A_2OA_3 > 60^{\circ}$ and so $O$ lies strictly inside the circumcircle of $ZA_2A_3$, where $Z$ is the centre of the regular hexagon. Similarily $O$ lies strictly inside the circumcircle of triangles $ZA_4A_5$ and $ZA_6A_1$, impossible since these three concur at $Z$. This final contradiction finishes the problem.

Assume it is possible. By rotation and homothety, we can put the vertices of the equilateral triangle to $A~(1,0)$, $B~(-1,0)$ and $C~\left(0,\sqrt3\right)$. Then the vertices of the regular hexagon $V_1V_2V_3V_4V_5V_6$ (counterclockwise orientation) can only be in the three triangular regions $\mathcal A$, $\mathcal B$, $\mathcal C$, as shown below. [asy][asy] pair a,b,c;a=W;b=E;c=sqrt(3)*N; fill(a--(2a-b)--(2a-c)--cycle,green); fill(b--(2b-a)--(2b-c)--cycle,green); fill(c--(2c-b)--(2c-a)--cycle,green); draw((2b-a)--(2a-b)^^(2b-c)--(2c-b)^^(2a-c)--(2c-a)); label("$A$",a,NW);label("$B$",b,NE);label("$C$",c,E); label("$\mathcal A$",(5a-b-c)/3); label("$\mathcal B$",(5b-a-c)/3); label("$\mathcal C$",(5c-b-a)/3); [/asy][/asy] In the following, subscripts take modulo $6{}$. Claim0.There exists least one vertex $V_k$ ($1\le k\le6$) in each of $\mathcal A$, $\mathcal B$, $\mathcal C$. Assume otherwise and let there be no vertex in region $\mathcal A$. Then the hexagon $V_1V_2V_3V_4V_5V_6$ is entirely to the right of line $BC$ and thus $\triangle ABC{}$ isn't inside $V_1V_2V_3V_4V_5V_6 $. Contradiction. Claim1.Points $V_i~(x_1,y_1)$ and $V_{i+3}~(x_2,y_2)$ cannot be in the same region. Assume otherwise and let them both in $\mathcal C$. So $y_1\ge\sqrt3\pm\sqrt3x_1$, $y_2\ge\sqrt3\pm\sqrt3x_2$. By complex numbers, $V_{i+2}~\left(\tfrac{x_1+3x_2-\sqrt3y_1+\sqrt3y_2}4,\tfrac{\sqrt3x_1-\sqrt3x_2+y_1+3y_2}4\right)$. But \[\sqrt3x_1-\sqrt3x_2+y_1+3y_2>y_1+\sqrt3x_1-\sqrt3+y_2-\sqrt3x_2-\sqrt3\ge0.\]So $V_{i+2}$ is above the line $AB{}$ and therefore can only be in $\mathcal C$. We can similarly prove that each of $V_1$, $V_2$, $V_3$, $V_4$, $V_5$, $V_6$ is inside $\mathcal C$ and using Claim 0 yields Contradiction. Claim2.If $V_i~(x_1,y_1)$ is in the same region as $V_{i+2}~(x_2,y_2)$, then so are $V_{i+1}$ and $V_{i+4}$. Let $V_i$ and $V_{i+2}$ be both in $\mathcal C$. So $y_1\ge\sqrt3\pm\sqrt3x_1$, $y_2\ge\sqrt3\pm\sqrt3x_2$. By complex numbers, $V_{i+1}~\left(\tfrac{\sqrt3x_1+\sqrt3x_2-y_1+y_2}{2\sqrt3},\tfrac{x_1-x_2+\sqrt3y_1+\sqrt3y_2}{2\sqrt3}\right)$. Since \[x_1-x_2+\sqrt3y_1+\sqrt3y_2>\frac{y_1+\sqrt3x_1-\sqrt3}{\sqrt3}+\frac{y_2-\sqrt3x_2-\sqrt3}{\sqrt3}\ge0,\]$V_{i+1}$ is above the line $AB{}$ and therefore can only be in $\mathcal C$. Additionally, $V_{i+4}~\left(\tfrac{x_1+x_2+\sqrt3y_1-\sqrt3y_2}2,\tfrac{y_1+y_2-\sqrt3x_1+\sqrt3x_2}2\right)$, and since \[y_1+y_2-\sqrt3x_1+\sqrt3x_2>y_1-\sqrt3x_1-\sqrt3+y_2+\sqrt3x_2-\sqrt3\ge0,\]$V_{i+4}$ is above the line $AB{}$ and therefore can only be in $\mathcal C$. Consider the multiset \[S=\big\{\#\{1\le i\le6:V_i\in\mathcal A\}, \#\{1\le i\le6:V_i\in\mathcal B\}, \#\{1\le i\le6:V_i\in\mathcal C\}\big\}.\]Each element of $S{}$ has to be greater than or equal to one (by Claim 0) and less than or equal to three (otherwise by pigeon hole principle there exists a pair of opposite vertices both chosen and this yields contradiction with Claim 1). We therefore have the following possibilities: $\{1,2,3\}$; $\{2,2,2\}$. For the first, Claim2 says that the "$2{}$" must be adjacent, say $V_1$ and $V_2$. But by Claim2, the pairs $(V_4,V_6)$ and $(V_3,V_5)$ can't be in the same region otherwise equilateral $\triangle V_1V-3V_5$ or $V_2V_4V_6$ yields contradiction with Claim2. So the only possibility is $S=\{2,2,2\}$, from which the solution is similar to StarChan's.