I am starting to study fundamental groups and came across the following result:
Let be $\;X=\big\{(x,y)\in\mathbb{R}^2: 2x^2-11xy-2y^2=0 \big\}\subset\mathbb{R}^2$.
I Have to calculate the $\,\pi_1(X)\,$ of a set $X$.
I solved the second grade equation by the y and I created the 3 sets :
$\;X_1=\big\{(x,y)\in\mathbb{R}^2: y=-\frac{(11-\sqrt{137})}{4}x\}\subset\mathbb{R}^2$ $\bigcup$ $\big\{(x,y)\in\mathbb{R}^2: y=-\frac{(11+\sqrt{137})}{4}x,x>0\}\subset\mathbb{R}^2$.
$\;X_2=\big\{(x,y)\in\mathbb{R}^2: y=-\frac{(11+\sqrt{137})}{4}x\}\subset\mathbb{R}^2$ .
$\;X_1\cap\;X_2 =\big\{(x,y)\in\mathbb{R}^2: y=-\frac{(11-\sqrt{137})}{4}x, x>0\}\subset\mathbb{R}^2$.
We can use the Saifert-Van Kampen Theorem to say that $\,\pi_1(X_1)\,$ is trivial (because $\;X_1$ is contractible to a point) ,the same for $\,\pi_1(X_2)\,$ and $\,\pi_1(X_2\cap X_1)\,$ .
Then $\,\pi_1(X) \simeq \dfrac{ \pi_1(X_1)\ast\pi_1(X_2) } {\mathit{H}} = \dfrac{\{\emptyset\}\ast\{\emptyset\}}{\{\emptyset\}}=\{\emptyset\} ,$
I do not know if my solution is right. Can someone tell me if is it the right answer?
Your proof is correct. However, in this case there is an easier way of doing it; namely, by showing your space is actually contractible.
Indeed, note $$2x^2-11xy-y^2=2\left(\left(x-\frac{11}{4}y\right)^2-\left(\frac{\sqrt{137}}{4}y\right)^2\right).$$
Therefore, if you consider the homeomorphism $h: \mathbb{R}^2 \to \mathbb{R}^2; h(x,y)=\left(x-\frac{11}{4}y,\frac{\sqrt{137}}{4}y\right)$, you get $$X \cong h(X)=\{(x,y) \in \mathbb{R}^2 \ | \ x^2-y^2=0\} = \{(x,y) \in \mathbb{R}^2 \ | \ x = y \} \cup \{(x,y) \in \mathbb{R}^2 \ | \ x = -y \}.$$ Therefore, $X$ is homeomorphic to the union of the diagonals of the coordinate plane. The latter is clearly contractible, and hence, so is $X$. In particular, $\pi_1(X,\cdot)=\{1\}$.
Remark. Be careful! There are no empty groups (they all have an identity). A trivial group is the group with only one element; namely, the identity.