Maximum and minimum value of $x^2+y^2-2x$ for $21x^2-6xy+29y^2+6x-58y-151=0$
What I tried is:
The Lagrange multiplier theorem.
$$\Lambda (x,y,\lambda)=f(x,y)+\lambda g(x,y)=x^2+y^2-2x+\lambda(21x^2-6xy+29y^2+6x-58y-151)$$ then solve the following system of equation $$\left\{\begin{matrix} \cfrac{d\Lambda}{dx}=0&&&&&&(1)\\ \cfrac{d\Lambda}{dy}=0&&&&&&(2)\\ \cfrac{d\Lambda}{d\lambda}=0&&&&&&(3)\\ \end{matrix} \right.$$
Can I solve it without Lagrange method? Help me please.
Yes, this can be done without Lagrange's method. Let us define, $ f(x,y) := 21x^2-6xy+29y^2+6x-58y-151=0$ to be your constraint. Further, let us change the coordinates to $X:=x-1$, $Y:=y$. Then your optimization problem is to find the minimum and maximum of $X^2 + Y^2 - 1$ or equivalently, of $X^2 + Y^2$ such that $f(X+1, Y)=0$. The key here is to note that $f(x,y)$ (and hence $f(X+1, Y)=0$) is a shifted ellipse (the constraint is of the form of a conic section, and you can prove that this is an ellipse; alternatively, see Wolfram Alpha), and we are looking for the point on this ellipse which is closest and furthest away from the origin. This can be done through various means.
In particular, you can look for points on the ellipse such that the normal through those points passes through the origin (this is equivalent to finding a circle through the origin which is tangential to the ellipse).
Or you can also parameterize your ellipse as has been done here.