Well, yes, "Lagrange multipliers" is a Calc 3 type problem. You said "for Lagrange Multiplier". Do you know what that means?
The problem asks you to find those numbers, x and y, out of those that satisfy 2x+ 3y= 6, that give minimum values for x^2+ y^2. Note that 2x+3y= 6 is the straight line through (3, 0) and (0, 2) while x^2+ y^2= r^2 with center at (0, 0) and radius r.
But you don't need "Lagrange multipliers" for this problem- you really just need secondary school geometry and maybe a little "PreCalculus".
There are no such numbers that make x^2+ y^2 a maximum. We can have arbitrarily large circles that pass through the line 2x+ 3y= 6 so there exist arbitrarily large |x| and |y| on that line that will make x^2+ y^2 arbitrarily large.
However, a very small circle may miss that line altogether. Geometrically, the circle with smallest radius that includes any point on that line is tangent to the line. The line itself, which we can write as y= 2- (2/3)x, has slope -2/3. The line through the origin (the center of the circle) perpendicular to that has equation y= (3/2)x. The point where those two line intersect is the point on the circle x^2+ y^2= r^2 that minimizes r^2.
The "Lagrange multiplier" method says that points satifying the "constraint" g(x,y)= constant, that minimize or maximize f(x,y), must have gradient vectors parallel: \nabla f(x,y)= \lambda \nabla g(x,y) where the constant, \lambda, is the "Lagrange multiplier".
Here, f(x,y)= x^2+ y^2 so \nabla f(x,y)= 2x\vec{i}+ 2y\vec{j} and g(x,y)= 2x+ 3y so \nabal g(x,y)= 2\vec{i}+ 3\vec{j}. \nabla f(x,y)= \lambda \nable g(x,y) becomes 2x\vec{i}+ 2y\vec{j}= \lambda(2y\vec{i}+ 3\vec{j} which gives the two equations 2x= 2\lambda and 2y= 3\lambda.
You can eliminate "\lambda", which is not really part of the solution, by dividing one equation by the other. That will give you one equation in x and y. Remember that they must also satisfy 2x+ 3y= 6.