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**Hint:**In these types of question remember that slope of tangent is given as $ \dfrac{{dy}}{{dx}} $ and also remember that equation of tangent is given as $ \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right) $ here m is the slope and x and y are the passing point, use this information to approach the solution.

**Complete step-by-step answer:**

The given equation is $ y = \dfrac{1}{{{x^2} - 2x + 3}} $

We know that slope of tangent is given by $ \dfrac{{dy}}{{dx}} $

So, differentiating the given equation with respect to x we get,

\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}{\left( {{x^2} - 2x + 3} \right)^{ - 1}}\]

We know that \[\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}\]

Therefore, \[\dfrac{{dy}}{{dx}} = - 1{\left( {{x^2} - 2x + 3} \right)^{ - 2}}\dfrac{d}{{dx}}\left( {{x^2} - 2x + 3} \right)\]

$ \Rightarrow $ \[\dfrac{{dy}}{{dx}} = - 1{({x^2} - 2x + 3)^{ - 2}}(2x - 2)\]

$ \Rightarrow $ \[\dfrac{{dy}}{{dx}} = \dfrac{{ - 2(x - 1)}}{{{{({x^2} - 2x + 3)}^2}}}\]

Since the slope given 0

Therefore, \[\dfrac{{dy}}{{dx}} = 0\]

\[\dfrac{{dy}}{{dx}} = \dfrac{{ - 2(x - 1)}}{{{x^2} - 2x + 3}} = 0\]

$ \Rightarrow $ \[ - 2(x - 1) = 0\]

$ \Rightarrow $ \[x - 1 = 0\]

$ \Rightarrow $ x = 1

We got the value of x so for the value of y substituting the value of x in the given equation i.e. $ y = \dfrac{1}{{{x^2} - 2x + 3}} $ we get

$ \Rightarrow $ \[y = \dfrac{1}{{{1^2} - 2(1) + 3}} = \dfrac{1}{2}\]

$ \Rightarrow $ \[y = \dfrac{1}{2}\]

Now we know that the tangent whose slope is zero passes through point $ \left( {1,\dfrac{1}{2}} \right) $

We know that when a equation of tangent passing through (x, y) which have slope m is given as $ \left( {y - {y_1}} \right) = m\left( {x - {x_1}} \right) $

So, the equation of the curve passes through the point $ \left( {1,\dfrac{1}{2}} \right) $ whose slope is zero i.e. m = 0

$ \left( {y - \dfrac{1}{2}} \right) = 0(x - 1) $

$ \Rightarrow $ $ y - \dfrac{1}{2} = 0 $

Hence the required equation is $ y = \dfrac{1}{2} $ .

**Note:**In these types of question of finding equation of tangent we have to obtain the passing point through which tangent is passing which we can find using the method of differentiation and since we had given the slope equal to zero from that we can find the value of x and y and substitute the values in the equation of tangent.

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