1. Use the tangent-intercept condition to find $k$.

Let $P(x,y)$ be any point on the curve, and let the tangent at $P$ meet the positive axes at $A(a,0), \quad B(0,b).$ So the tangent is in intercept form: $\frac{X}{a}+\frac{Y}{b}=1.$ Since $P(x,y)$ lies on this tangent, \frac{x}{a}+\frac{y}{b}=1. \tag{1}

Now given $PA:PB=1:k$.

For a point dividing the segment $AB$ internally in the ratio $1:k$, we have $P=\left(\frac{k a}{k+1},\frac{b}{k+1}\right).$ Hence $x=\frac{k a}{k+1}, \qquad y=\frac{b}{k+1}.$ So $a=\frac{(k+1)x}{k}, \qquad b=(k+1)y.$ Substitute into (1): $\frac{x}{(k+1)x/k}+\frac{y}{(k+1)y}=\frac{k}{k+1}+\frac{1}{k+1}=1,$ which is consistent.

The slope of the tangent in intercept form is $m=-\frac{b}{a}=-\frac{(k+1)y}{(k+1)x/k}=-\frac{k y}{x}.$ Thus the differential equation of the curve is \frac{dy}{dx}=-\frac{k y}{x}. \tag{2}

2. Solve the curve equation using the two given points.

From (2): $\frac{dy}{y}=-k\frac{dx}{x}.$ Integrating, $\ln y=-k\ln x+C$ so y=Cx^{-k}. \tag{3}

Since the curve passes through $(1,1)$, $1=C\cdot 1^{-k} \implies C=1.$ Hence $y=x^{-k}.$

Now use the point $\left(\frac{1}{10},100\right)$: $100=\left(\frac{1}{10}\right)^{-k}=10^k.$ Thus $10^k=10^2 \implies k=2.$

3. Solve the given differential equation.

We are given $e^{dy/dx}=kx+\frac{k}{2}, \qquad y(0)=k.$ Since $k=2$, $e^{dy/dx}=2x+1.$ Taking natural log, $\frac{dy}{dx}=\ln(2x+1).$ Integrate: $y=\int \ln(2x+1)\,dx + C.$

Let $u=2x+1$, so $du=2dx$, $dx=\frac{du}{2}$: