The question is the same as the task of finding the minimum of the function $f(x,y)=\sqrt{x^2+y^2}$, subject to the constraint $$x+2y-3=0\text.$$ We can as well set the task of finding the root of the minimum of the function $F=f^2$ under the given side condition. This is easier because we get, at first, rid of the square root function.
We discuss two methods:

1. Direct method

In the constraint, we can isolate the variable $y$: $$y=-{{1}\over{2}}x+{{3}\over{2}}\text.$$ Substitution in $F(x,y)=f(x,y)^2$, gives the formula $$x^2+\left(-{{1}\over{2}}x+{{3}\over{2}}\right)^2\text.$$ This valley parabola has a minimum in $x$ that meets the requirement $$2x+2\cdot \left(-{{1}\over{2}}x+{{3}\over{2}}\right)\cdot -{{1}\over{2}}=0$$ that is $${{5}\over{2}}x-{{3}\over{2}}=0\text.$$ Thus: $$x={{3}\over{5}}\text.$$ The value of $y$ follows from the constraint equation written as $$y=-{{1}\over{2}}x+{{3}\over{2}}$$ and is thus equal to $$y=-{{1}\over{2}}\cdot {{3}\over{5}}+{{3}\over{2}}={{6}\over{5}}$$ For the values of $x$ and $y$ found the function value $f(x,y)$ is minimal and is equal to $$\sqrt{\left({{3}\over{5}}\right)^2+\left({{6}\over{5}}\right)^2}=\frac{3}{5}\sqrt{5}\text.$$

2. Lagrange multipliers

The task to find the minimum of

$$F(x,y)=x^2+y^2$$ subject to the constraint $$G(x,y)=0$$ with $$G(x,y)=x+2y-3$$ is in the method of Lagrange multipliers equivalent to the task of finding the critical point of the function $$H(x,y,\lambda) = F(x,y) + \lambda \cdot G(x,y)$$ in the three variables $x$, $y$ and $\lambda$. You have to solve the following system of equations in three variables: $$\left\{\frac{\partial H}{\partial x}=0,\quad \frac{\partial H}{\partial y}=0,\quad \frac{\partial H}{\partial \lambda}=0\right\}$$ in this case we have $$H(x,y,\lambda) = x^2+y^2 +\lambda\left(x+2y-3\right)$$ and the system is equal to $$\left\{ \lambda+2 x=0 ,\quad 2 \lambda+2 y=0 ,\quad x+2 y- 3=0 \right\}$$ The solution of this system is $$\left\{ x={{3}\over{5}} ,\quad y={{6}\over{5}} ,\quad \lambda=-{{6 }\over{5}} \right\}$$ Once again we find the same values $x$ and $y$ for which $F(x,y)$ is minimal under the given condition. This way you also find the values $x$ and $y$ for which $f(x,y)$ is minimal; this minimal value equals $$\sqrt{\left({{3}\over{5}}\right)^2+\left({{6}\over{5}}\right)^2}=\frac{3}{5}\sqrt{5}\text.$$

The method of Lagrange multipliers appeared out of the blue, but could be better understood as follows.

The constraint equation $G(x,y)=0$ represents a straight line. We can now consider the contour curves of the function $F$: these are circles with the origin as centre. If a contour curve intersects the straight line in two points, then the corresponding level of $F$ cannot be an extreme value, because a slightly larger or slightly smaller value leads to curves that intersect the line with equation $G(x,y)=0$ in two points, too. An extremum (in this case a minimum) can only occur if the straight line and the contour curve touch each other, i.e., when the line is a tangent line of the contour curve. But this means that we must find a point $(a,b)$ for which $F(a,b)=G(a,b)$ and for which the gradient of $F$ and the gradient of $G$ calculated in that point are a multiple of each other. The requirements

$$F(a,b)=G(a,b)\qquad\text{and}\qquad {\nabla}F(a,b)= -\lambda\cdot {\nabla}G(a,b)\quad\text{for some number }\lambda\small,$$ for finding the point $(a,b)$ correspond to the task to solve the system of equations $$\left\{\frac{\partial H}{\partial \lambda}=0,\quad \frac{\partial H}{\partial x}=0,\quad \frac{\partial H}{\partial y}=0\right\}$$ for $$H(x,y,\lambda) = F(x,y) + \lambda \cdot G(x,y)\tiny.$$ The extreme value is equal to $F(a,b)$ for the solution $(a,b)$ of the system of equations.