Differentiation on tangent on a point

Let $y=e^{x^2}+ln x+e^{xln x}$.

Using the fact that $e^{ln x}=x$, we have $$y=e^{x^2}+ln x+x^x$$

Let us suppose for simplicity, $u=e^{x^2}$, $v=ln x$ and $w=x^x$. $$y=u+v+w$$ $$implies frac{dy}{dx}=frac{du}{dx}+frac{dv}{dx}+frac{dw}{dx}quad (*)$$

Differentiating $u$

To differentiate $u=e^{x^2}$, we let $t=x^2$. By Chain Rule, $$frac{du}{dx}=frac{du}{dt}cdot frac{dt}{dx}$$ $$frac{du}{dx}=frac{d}{dt}(e^t)cdot frac{d}{dx}(x^2)$$ $$frac{du}{dx}=2xe^{x^2}quad (1)$$

Differentiating $v$ is trivial as we know $(ln x)'=1/x$. So, $$frac{dv}{dx}=frac{1}{x}quad (2)$$

Differentiating $w$

For this, take $ln$ of both sides $$ln w =xln x$$ Differentiate both sides w.r.t. $x$, $$frac{1}{w}cdotfrac{dw}{dx}=xcdotfrac{1}{x}+ln xcdot 1$$ $$frac{1}{x^x}cdot frac{dw}{dx}=1+ln x$$ $$frac{dw}{dx}=x^x(1+ln x)quad (3)$$

Using $(1)$, $(2)$ and $(3)$ in $(*)$, we get $$frac{dy}{dx}=2xe^{x^2}+frac{1}{x}+x^x(1+ln x)$$

Thus the slope of $y$ at the point $(1,e+1)$ is obtained by putting $x=1$, $$left[frac{dy}{dx}right]_{x=1}=2e+1+1=2e+2$$

Thus, the tangent has slope $m=2e+2$ and lies on $(x_0,y_0)=(1, e+1)$. By point-slope form, $$y-y_0=m(x-x_0)$$ So, the equation of required tangent is $$y-e-1=(2e+2)(x-1)$$ $$(2e+2)x-y-(e-1)=0$$

Hope this helps :)