The tangent to the curve, $$y = xe^{x^2}$$ passing through the point $$(1, e)$$ also passes through the point:

We are given the curve $$y = x e^{x^{2}}$$ and the point $$(1,e)$$ which definitely lies on this curve because

$$y(1)=1 \cdot e^{1^{2}} = 1 \cdot e = e.$$

We have to find the tangent to the curve at this point and then check which of the four listed points also lies on this tangent.

To get the tangent we first need its slope. For that, we differentiate the curve. We recall the product rule: if $$u(x)$$ and $$v(x)$$ are functions of $$x$$, then

$$\frac{d}{dx}\,[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).$$

Here $$u(x)=x$$ and $$v(x)=e^{x^{2}}.$$ So,

$$\frac{dy}{dx}=u'(x)v(x)+u(x)v'(x).$$

Now $$u'(x)=1$$ and $$v'(x)=e^{x^{2}}\cdot 2x$$ (because the derivative of $$e^{x^{2}}$$ is $$e^{x^{2}}\cdot 2x$$ by the chain rule).

Substituting these derivatives we get

$$\frac{dy}{dx}=1\cdot e^{x^{2}} + x\cdot (e^{x^{2}}\cdot 2x)=e^{x^{2}} + 2x^{2}e^{x^{2}}.$$

We can factor out $$e^{x^{2}}$$ to write

$$\frac{dy}{dx}=e^{x^{2}}\bigl(1+2x^{2}\bigr).$$

We need the slope at $$x=1$$, so we substitute $$x=1$$:

$$m = \left.\frac{dy}{dx}\right|_{x=1}=e^{1^{2}}\bigl(1+2\cdot 1^{2}\bigr)=e\,(1+2)=3e.$$

Thus the slope of the tangent at $$(1,e)$$ is $$3e.$$

The point-slope form of a straight-line equation is

$$y - y_{1}=m(x - x_{1}),$$

where $$(x_{1},y_{1})=(1,e)$$ and $$m=3e.$$ Substituting these, we get

$$y - e = 3e\,(x - 1).$$

Now we simplify step by step:

First expand the right side:

$$y - e = 3e\,x - 3e.$$

Add $$e$$ to both sides:

$$y = 3e\,x - 3e + e.$$

Simplify the constants:

$$y = 3e\,x - 2e.$$

Factor out $$e$$ to make the relationship clearer:

$$y = e\,(3x - 2).$$

Any point $$(x,y)$$ lying on this tangent must satisfy $$y = e\,(3x - 2).$$ We now test each option.

Option A: $$\left(\dfrac{4}{3},\,2e\right).$$ Substitute $$x=\dfrac{4}{3}$$ in the right-hand side:

$$e\bigl(3\cdot \tfrac{4}{3}-2\bigr)=e\,(4-2)=2e.$$

The resulting $$y=2e$$ matches the given $$y$$-coordinate, so this point lies on the tangent.

Option B: $$(2,\,3e).$$ Put $$x=2$$:

$$e\,(3\cdot 2-2)=e\,(6-2)=4e\neq 3e,$$

so Option B is not on the line.

Option C: $$\left(\dfrac{5}{3},\,2e\right).$$ Put $$x=\dfrac{5}{3}:$$

$$e\bigl(3\cdot \tfrac{5}{3}-2\bigr)=e\,(5-2)=3e\neq 2e,$$

hence Option C is not on the line.

Option D: $$(3,\,6e).$$ Put $$x=3:$$

$$e\,(3\cdot 3-2)=e\,(9-2)=7e\neq 6e,$$

so Option D is also not on the line.

The only option that satisfies the tangent’s equation is Option A.

Hence, the correct answer is Option A.